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3

## How to solve this equation? 2cosx+1=0;x in[0,3pi);

Eddie
Featured 1 month ago

$\frac{2 \pi}{3} , \frac{4 \pi}{3} , \frac{8 \pi}{3}$

#### Explanation:

$2 \cos x + 1 = 0$

$\cos x = - \frac{1}{2}$

Now we know from this handy "CAST" nmemonic that cos is negative in Q2+Q3:

We can even draw the triangles:

So in Q1: $\cos x = \frac{1}{2} \implies x = \frac{\pi}{3}$

The corresponding angle in Q2 is $\pi - \frac{\pi}{3} = \frac{2 \pi}{3}$

The corresponding angle in Q3 is $\pi + \frac{\pi}{3} = \frac{4 \pi}{3}$

Now because it is $x \in \left[0 , 3 \pi\right)$, we are not done yet.

After spinning round a full $2 \pi$, we have another $\pi$ to go which takes us back into Q2.

The second corresponding angle in Q2 is $2 \pi + \left(\pi - \frac{\pi}{3}\right) = \frac{8 \pi}{3}$

1

## How do you sketch the graph of y=3 cos (x+π) -3?

HSBC244
Featured 1 month ago

I like getting rid of the phase shift (the $x + \pi$ part) using the sum and difference formulas. The one that is applicable here is

$\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$.

We have:

$y = 3 \left(\cos x \cos \left(\pi\right) - \sin x \sin \pi\right) - 3$

$y = 3 \left(\cos x \left(- 1\right) - 0\right) - 3$

$y = - 3 \cos x - 3$

Now you need a little bit of knowledge on the basic cosine function, $y = \cos x$. Here's the graph:

graph{y = cosx [-10, 10, -5, 5]}

Whenever there is a coefficient $a$ next to the cosine, you have an altered amplitude, which is the distance between the centre (the line $y = 0$) and the top or bottom of the curve.

In the graph of $y = \cos x$, the amplitude is simply $1$. In the graph of $y = - 3 \cos x - 3$, the amplitude will be $3$.

The $-$ is in front of the $3$ to signify a reflection over the x-axis.

Finally, the $- 3$ to the far right of the equation signifies a vertical transformation of $3$ units down. We are left with the following graph:

graph{y = -3cosx - 3 [-10, 10, -5, 5]}

Hopefully this helps!

1

## How to find amplitude and cycle for y = 1+4 sin (2x)?

Love K. Soother
Featured 1 month ago

Period of $y = 1 + 4 \setminus \sin \left(2 x\right)$ is $n \setminus \pi$ and amplitude is $4$

### GENERAL WAVE EQUATION

$f \left(x\right) = a \setminus \sin \left(b x + c\right) + d$,

where $\setminus \pm c$, $\setminus \pm d$ are wave right (left) or up (down) shifts along $x \text{-axis}$ and $y \text{-axis}$ respectively.

### SOLUTION (for period)

Ignoring the shifts along axes i.e. assuming $c = d = 0$ we can have our period and amplitude without any sort of error.

$f \left(x\right) = y = 1 + 4 \setminus \sin \left(2 x\right)$

$\setminus R i g h t a r r o w y - 1 = 4 \setminus \sin \left(2 x\right)$

$\setminus R i g h t a r r o w g \left(x\right) = 4 \setminus \sin \left(2 x\right)$

We know from fundamental trigonometric ideas that the period of sine function is $2 n \setminus \pi$ i.e.

$\setminus \sin \left(x\right) = \setminus \sin \left(2 n \setminus \pi + x\right)$,

Let,

$\setminus \sin \left(u\right) = \setminus \sin \left(2 n \setminus \pi + u\right)$

Then for $u = 2 x$, we have;

$\setminus \sin \left(2 x\right) = \setminus \sin \left(2 n \setminus \pi + 2 x\right)$

Multiplying both sides by $4$,

$4 \setminus \sin \left(2 x\right) = 4 \setminus \sin 2 \left(n \setminus \pi + x\right)$

$\setminus \equiv g \left(x\right) = g \left(n \setminus \pi + x\right)$

Hence period is $n \setminus \pi$

### SOLUTION (amplitude)

In general the amplitude of sine function is given as;

If $y = \setminus \sin \left(x\right)$ then amplitude is $1$ unit long. as followed by graph

graph{y=\sin(x)[-3,3,-1.5,1.5]}

Our equation is a sort of $y = a \sin \left(b x\right)$

Which becomes;

$y = a \left[\setminus \sin \left(b x\right)\right]$

And that expands the graph of ordinary sine function i.e. $\sin \left(x\right)$ up to $| a |$ times along $y \text{-axis}$

graph{y=4\sin(2x)}

Therefore our amplitude is $4$

The graph of our real function $y = 1 + 4 \setminus \sin \left(2 x\right)$ is shifted upwards $1$ unit

graph{y=1+4 \sin(2x)}

But amplitude remains same.

2

## Expressing in different form and finding a maximum? See picture below (7 marks)

dk_ch
Featured 1 month ago

(a)

$\sqrt{15} \sin \left(2 x\right) + \sqrt{5} \cos \left(2 x\right)$

$= \sqrt{20} \left(\frac{\sqrt{15}}{\sqrt{20}} \sin \left(2 x\right) + \frac{\sqrt{5}}{\sqrt{20}} \cos \left(2 x\right)\right)$

[Taking $\frac{\sqrt{15}}{\sqrt{20}} = \cos \alpha \mathmr{and} \frac{\sqrt{5}}{\sqrt{20}} = \sin \alpha$]

$= \sqrt{20} \left(\cos \alpha \sin \left(2 x\right) + \sin \alpha \cos \left(2 x\right)\right)$

$= 2 \sqrt{5} \sin \left(2 x + \alpha\right)$

Where $\alpha = {\tan}^{-} 1 \left(\frac{\sqrt{5}}{\sqrt{15}}\right) = {\tan}^{-} 1 \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

(b)

(i)
$f \left(x\right) = \frac{2}{5 + \sqrt{15} \sin \left(2 x\right) + \sqrt{\cos} \left(2 x\right)}$

$\implies f \left(x\right) = \frac{2}{5 + 2 \sqrt{5} \sin \left(2 x + \alpha\right)}$

The value of $f \left(x\right)$ will be maximum when $\sin \left(2 x + \alpha\right)$ is minimum i.e. $\sin \left(2 x + \alpha\right) = - 1$

So

$f {\left(x\right)}_{\text{max}} = \frac{2}{5 - 2 \sqrt{5}}$

$\implies f {\left(x\right)}_{\text{max}} = \frac{2 \left(5 + 2 \sqrt{5}\right)}{\left(5 - 2 \sqrt{5}\right) \left(5 + 2 \sqrt{5}\right)}$

$\implies f {\left(x\right)}_{\text{max}} = \frac{10 + 4 \sqrt{5}}{25 - 20} = 2 + \frac{4}{5} \sqrt{5}$

(ii)Here $f \left(x\right)$ is maximum when

$\sin \left(2 x + \alpha\right) = - 1$

$\implies \sin \left(2 x + \frac{\pi}{6}\right) = \sin \left(\frac{3 \pi}{2}\right)$

$\implies 2 x + \frac{\pi}{6} = \frac{3 \pi}{2}$

$\implies 2 x = \frac{3 \pi}{2} - \frac{\pi}{6} = \frac{4 \pi}{3}$

$\implies x = \frac{2 \pi}{3}$

2

## What is the frequency of f(t)= sin(4t) - cos(7t)?

Douglas K.
Featured yesterday

${f}_{0} = \frac{1}{2 \pi} \text{ Hz}$

#### Explanation:

Given: $f \left(t\right) = \sin \left(4 t\right) - \cos \left(7 t\right)$ where t is seconds.

Use this reference for Fundamental Frequency

Let ${f}_{0}$ be the fundamental frequency of the combined sinusoids, in Hz (or ${\text{s}}^{-} 1$).

${\omega}_{1} = 4 \text{ rad/s}$
${\omega}_{2} = 7 \text{ rad/s}$

Using the fact that $\omega = 2 \pi f$

${f}_{1} = \frac{4}{2 \pi} = \frac{2}{\pi} \text{ Hz}$ and ${f}_{2} = \frac{7}{2 \pi} \text{ Hz}$

The fundamental frequency is the greatest common divisor of the two frequencies:

${f}_{0} = \gcd \left(\frac{2}{\pi} \text{ Hz", 7/(2pi)" Hz}\right)$

${f}_{0} = \frac{1}{2 \pi} \text{ Hz}$

Here is a graph:

graph{y = sin(4x) - cos(7x) [-10, 10, -5, 5]}

Please observe that it repeats every $2 \pi$

2

## Express r=cos(2theta) in rectangular coordinates?

Douglas K.
Featured 1 month ago

${x}^{2} + {y}^{2} = {\left(\frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}\right)}^{2}$

#### Explanation:

Here is the graph of the original equation $r = \cos \left(2 \theta\right)$:

Use the identity $\cos \left(2 \theta\right) = {\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)$

$r = {\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)$:

Substitute the $\sqrt{{x}^{2} + {y}^{2}}$ for r, ${x}^{2} / \left({x}^{2} + {y}^{2}\right)$ for ${\cos}^{2} \left(\theta\right)$, and ${y}^{2} / \left({x}^{2} + {y}^{2}\right)$ for ${\sin}^{2} \left(\theta\right)$:

$\sqrt{{x}^{2} + {y}^{2}} = {x}^{2} / \left({x}^{2} + {y}^{2}\right) - {y}^{2} / \left({x}^{2} + {y}^{2}\right)$

$\sqrt{{x}^{2} + {y}^{2}} = \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}$

Here is the graph of that equation:

graph{sqrt(x^2+y^2)=x^2/(x^2+y^2)-y^2/(x^2+y^2) [-10, 10, -5, 5]}

It is only half of the loops

Square both sides:

${x}^{2} + {y}^{2} = {\left(\frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}\right)}^{2}$

Here is the graph of that equation:

graph{x^2+y^2=((x^2-y^2)/(x^2+y^2))^2 [-10, 10, -5, 5]}

This looks like the polar equation.

4

## PLZ help ! Challenge question of trig application!! with step plz?

Veerupaksh S.
Featured 4 weeks ago

$\textcolor{red}{{34}^{o}}$

#### Explanation:

We can construct this diagram from the given information.

Let length of $A B$ be ${l}_{1}$ and length of $B C$ be ${l}_{2}$.

It is given that ${l}_{1} / {l}_{2} = \frac{2}{3}$ $\to$ 1.

Also, observe that $B D = F E$ and $B F = D E$. $\to$ 2.

We have to find $\angle C A E$ which will give us elevation of C from A.

Let $\angle C A E$ be $\theta$.

From $\triangle A B F$ we observe that:-

$\frac{B F}{A B} = \sin {25}^{o} \implies B F = A B \cdot \sin {25}^{o} = 0.423 \cdot {l}_{1}$

$\therefore$ using 2., $D E = 0.423 \cdot {l}_{1}$ $\to$ 3.

$\frac{A F}{A B} = \cos {25}^{o} \implies A F = A B \cdot \cos {25}^{o} = 0.906 \cdot {l}_{1}$ $\to$ 4.

From $\triangle B C D$ we find:-

$\frac{C D}{B C} = \sin {40}^{o} \implies C D = B C \cdot \sin {40}^{o} = 0.643 \cdot {l}_{2}$ $\to$ 5.

$\frac{B D}{B C} = \cos {40}^{0} \implies B D = B C \cdot \cos {40}^{o} = 0.766 \cdot {l}_{2}$

$\therefore$ form 2., $F E = 0.766 \cdot {l}_{2}$ $\to$ 6.

Now,
$\tan \theta = \frac{C E}{A E} = \frac{C D + D E}{A F + F E}$

Substituting values of $D E , A F , C D , F E$ from 3., 4., 5., 6.

$\tan \theta = \frac{0.643 \cdot {l}_{2} + 0.423 \cdot {l}_{1}}{0.766 \cdot {l}_{2} + 0.906 \cdot {l}_{1}}$

Dividing the numerator and deniminator by ${l}_{2}$

$\implies \tan \theta = \frac{0.643 \cdot \cancel{{l}_{2} / {l}_{2}} + 0.423 \cdot {l}_{1} / {l}_{2}}{0.766 \cdot \cancel{{l}_{2} / {l}_{2}} + 0.906 \cdot {l}_{1} / {l}_{2}}$

Substituting value of ${l}_{1} / {l}_{2}$ from 1.

$\implies \tan \theta = \frac{0.643 + 0.423 \cdot \frac{2}{3}}{0.766 + 0.906 \cdot \frac{2}{3}} = 0.675$

$\therefore \theta = {\tan}^{-} 1 0.675 \approx {34}^{o}$

$\therefore \textcolor{red}{\angle C A E = {34}^{o}}$ which is the angle of elevation of $C$ from $A$.

2

## How do you convert 8=(3x-y)^2+2y+6x into polar form?

Douglas K.
Featured yesterday

#### Explanation:

Given: $8 = {\left(3 x - y\right)}^{2} + 2 y + 6 x$

Here is a graph of the Cartesian equation:

Expand the square:

$9 {x}^{2} - 6 x y + {y}^{2} + 6 x + 2 y - 8 = 0$

Substitute $r \cos \left(\theta\right)$ for every x:

$9 {\left(r \cos \left(\theta\right)\right)}^{2} - 6 \left(r \cos \left(\theta\right)\right) y + {y}^{2} + 6 \left(r \cos \left(\theta\right)\right) + 2 y - 8 = 0$

Substitute $r \sin \left(\theta\right)$ for every y:

$9 {\left(r \cos \left(\theta\right)\right)}^{2} - 6 \left(r \cos \left(\theta\right)\right) \left(r \sin \left(\theta\right)\right) + {\left(r \sin \left(\theta\right)\right)}^{2} + 6 \left(r \cos \left(\theta\right)\right) + 2 \left(r \sin \left(\theta\right)\right) - 8 = 0$

There is a common factor of ${r}^{2}$ in the first 3 terms:

$\left(9 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + {\sin}^{2} \left(\theta\right)\right) {r}^{2} + 6 \left(r \cos \left(\theta\right)\right) + 2 \left(r \sin \left(\theta\right)\right) - 8 = 0$

There is a common factor of r in the next 2 terms:

$\left(9 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + {\sin}^{2} \left(\theta\right)\right) {r}^{2} + \left(6 \cos \left(\theta\right) + 2 \sin \left(\theta\right)\right) r - 8 = 0$

The above is a quadratic equation in standard form where:

r in the independent variable

$a = 9 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + {\sin}^{2} \left(\theta\right)$

$b = 6 \cos \left(\theta\right) + 2 \sin \left(\theta\right)$

$c = - 8$

We use the identity cos^2(theta) + sin^2(theta) = 1 to simplify a:

$a = 8 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + 1$

We can use the quadratic formula to write r and a function of $\theta$ but, because radii should only be positive we will discard the negative root:

$r = \frac{- b + \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$

Substitute for every a:

$r = \frac{- b + \sqrt{{b}^{2} - 4 \left(8 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + 1\right) \left(c\right)}}{2 \left(8 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + 1\right)}$

Substitute for c:

$r = \frac{- b + \sqrt{{b}^{2} + 32 \left(8 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + 1\right)}}{2 \left(8 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + 1\right)}$

Substitute for every b:

$r = \frac{- \left(6 \cos \left(\theta\right) + 2 \sin \left(\theta\right)\right) + \sqrt{{\left(6 \cos \left(\theta\right) + 2 \sin \left(\theta\right)\right)}^{2} + 32 \left(8 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + 1\right)}}{2 \left(8 {\cos}^{2} \left(\theta\right) - 6 \cos \left(\theta\right) \sin \left(\theta\right) + 1\right)}$

Here is a graph of the polar equation:

The graphs are identical, therefore, the conversion is complete.

Note: I am sure that the expression under the radical can be simplified but I will leave that to you.

2

## Trig math problem help?

dk_ch
Featured yesterday

The given function representing the height $h$ in cm of the person above the ground with time $t$ is given by

$h = 15 \cos \left(\frac{2 \pi}{3} t\right) + 65$

(a) h is a cosine function of t, So it will have maximum value when $\cos \left(\frac{2 \pi}{3} t\right) = 1 = \cos 0 \implies t = 0$ s
and the maximum height of the swing becomes

${h}_{\max} = 15 \cos \left(\frac{2 \pi}{3} \times 0\right) + 65 = 80 c m$

(b) it takes $t = 0$sec to reach maximum height after the person starts.

c) The minimum height of the swing will be achieved when $\cos \left(\frac{2 \pi}{3} t\right) = - 1$

Minimum height

${h}_{\min} = 15 \times \left(- 1\right) + 65 = 50 c m$

(d). Again $\cos \left(\frac{2 \pi}{3} t\right) = - 1 = \cos \left(\pi\right) \implies t = \frac{3}{2} = 1.5$s

Here $t = 1.5$s represents the minimum time required to achieve the minimum height after start.

e) For $h = 60$ the equation becomes

$60 = 15 \cos \left(\frac{2 \pi}{3} t\right) + 65$

$\implies 60 - 65 = 15 \cos \left(\frac{2 \pi}{3} t\right)$

$\implies \cos \left(\frac{2 \pi}{3} t\right) = - \frac{1}{3}$

$\implies t = \left[\frac{1}{120} \left(2 n \times 180 \pm {\cos}^{-} 1 \left(- \frac{1}{3}\right)\right)\right] \sec \text{ where } n \in \mathbb{Z}$

$\implies t = \left(3 n \pm 0.912\right)$sec

when $n = 0 , t = 0.912 s$ this is the minimum time to reach at height $h = 60 c m$ after start. just after that height $h = 60 c m$ will be achieved again for $n = 1$

$\implies t = \left(3 \times 1 - 0.912\right) = 2.088$sec

So in $\left(2.088 - 0.912\right) = 1.176 s$ sec (the minimum time difference) within one cycle the swing will be less than 60 cm above the ground.

f) The height of the swing at 10 s can be had by inserting $t = 10$s in the given equation

${h}_{10} = 15 \cos \left(\frac{2 \pi}{3} \times 10\right) + 65 = 57.5$cm

2

## How do you use the quadratic formula to solve 12sin^2x-13sinx+3=0 in the interval [0,2pi)#?

Andy Y.
Featured yesterday

$x \approx 0.3398$, $x \approx 0.8481$, $x = 2.294$, and $x = 2.802$ radians.

#### Explanation:

Strategy: Rewrite this equation as a quadratic equation using $u = \sin \left(x\right)$. Solve the quadratic equation for $u$ by factoring. Then replace $u$ with $\sin \left(x\right)$ again and solve using $\arcsin$.

Step 1. Rewrite this equation as a quadratic using $\textcolor{red}{u} = \textcolor{red}{\sin} \left(x\right)$

You are given
$12 {\left(\textcolor{red}{\sin \left(x\right)}\right)}^{2} - 13 \left(\textcolor{red}{\sin \left(x\right)}\right) + 3 = 0$

Replace $\textcolor{red}{\sin \left(x\right)}$ with $\textcolor{red}{u}$
$12 {\textcolor{red}{u}}^{2} - 13 \textcolor{red}{u} + 3 = 0$

Step 2. Factor the quadratic equation.
$\left(3 u - 1\right) \left(4 u - 3\right) = 0$

Solving gives us
$3 u - 1 = 0$ and $4 u - 3 = 0$
$u = \frac{1}{3}$ and $u = \frac{3}{4}$

Step 3. Replace $u$ with $\sin \left(x\right)$ again and solve with $\arcsin$
$\sin \left(x\right) = \frac{1}{3}$ and $\sin \left(x\right) = \frac{3}{4}$

${\sin}^{- 1} \left(\sin \left(x\right)\right) = {\sin}^{- 1} \left(\frac{1}{3}\right)$ and ${\sin}^{- 1} \left(\sin \left(x\right)\right) = {\sin}^{- 1} \left(\frac{3}{4}\right)$

$x \approx 0.3398$ radians and $x \approx 0.8481$ radians

These answer work because we were asked to find the solutions in $\left[0 , 2 \pi\right] \approx \left[0 , 6.2832\right]$

However, the graph of $y = 12 {\left(\sin \left(x\right)\right)}^{2} - 13 \left(\sin \left(x\right)\right) + 3$ is

Which shows four solutions in the interval $\left[0 , 2 \pi\right]$, not two.

We must find the other solutions by recognizing that sine functions are periodic with respect to $\pi$

So, $x = \pi - 0.8481 = 2.294$ and $x = \pi - 0.3398 = 2.802$

The other two solutions are
$x = 2.294$ and $x = 2.802$