# Solve cosx+sinx=2/3 ?

Oct 4, 2016

Assuming you meant $\sin$($x$) $+$ $\cos$($x$) $=$ $\frac{2}{3}$,

$\cos \left({x}_{1}\right) = 0.96$
$\cos \left({x}_{2}\right) = - 0.29$ (to 2dp.)

#### Explanation:

To solve this equation, we have to use trigonometric functions to isolate one of the unknowns and solve for that unknown.

In this case, we will use ${\sin}^{2}$($x$) $+$ ${\cos}^{2}$($x$) = 1 and 2$\sin \left(x\right)$$\cos \left(x\right)$ = $\sin \left(2 x\right)$.

$\sin$($x$) $+$ $\cos$($x$) = $\frac{2}{3}$

${\left(\sin \left(x\right) + \cos \left(x\right)\right)}^{2}$ $=$ ${\left(\frac{2}{3}\right)}^{2} =$ $\frac{4}{9}$

${\sin}^{2} \left(x\right)$ $+$ ${\cos}^{2} \left(x\right)$ $+$ 2$\sin \left(x\right)$$\cos \left(x\right)$ $=$ $\frac{4}{9}$

$1$ $+$ $\sin \left(2 x\right)$ $=$ $\frac{4}{9}$

$\sin \left(2 x\right)$ $=$ $- \frac{5}{9}$

The function $\sin \left(x\right)$ is negative in the 3rd and 4th quadrants and $\sin \left(\theta\right) = \sin \left({180}^{\text{o}} - \theta\right)$. Therefore, we have two solutions for $2 x$:

$2 {x}_{1}$ $=$ ${\sin}^{-} 1 \left(- \frac{5}{9}\right)$ $=$ $- {33.75}^{\text{o}}$
$2 {x}_{2} = {180}^{\text{o"-(-33.75^"o") = 213.75^"o}}$

From this:
${x}_{1}$ $=$ $- {33.75}^{\text{o}} / 2$ $=$ $- {16.87}^{\text{o}}$
${x}_{2} = {213.75}^{\text{o"/2 = 106.87^"o}}$

$\cos \left({x}_{1}\right) = \cos \left(- {16.87}^{\text{o}}\right)$ $=$ $0.96$
$\cos \left({x}_{2}\right) = \cos \left({106.87}^{\text{o}}\right) = - 0.29$

Oct 5, 2016

$x = 1.87$-Round to 2 decimal places

#### Explanation:

$\sin x + \cos x = \frac{2}{3}$

Let's first use linear combination of cosine and sine with equal arguments formula to simplify it.

That is, we want to express
A cos x +B sin x in the form C cos (x-D). Note that A is the coefficient of cos x and B is the coefficient of sine x.

To find C use pythagorean formula and to find D we use one of these two formulas
$\cos D = \frac{A}{C} , \sin D = \frac{B}{C}$

From the given equation A = 1 and B = 1. So let's find C using pythagorean theorem.
$C = \sqrt{{A}^{2} + {B}^{2}} = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

To find D we need to first figure out which quadrant x is in and because both cos x and sin x are positive it means that x is in quadrant one.

$\cos D = \frac{1}{\sqrt{2}}$
$D = {\cos}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$

Therefore $\sin x + \cos x = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right)$

Now let's use that to solve the problem. That is,

$\sqrt{2} \cos \left(x - \frac{\pi}{4}\right) = \frac{2}{3}$

$\cos \left(x - \frac{\pi}{4}\right) = \frac{2}{3 \sqrt{2}}$

$x - \frac{\pi}{4} = {\cos}^{-} 1 \left(\frac{2}{3 \sqrt{2}}\right)$

$x = {\cos}^{-} 1 \left(\frac{2}{3 \sqrt{2}}\right) + \frac{\pi}{4}$

$x = 1.87$-Round to 2 decimal places

Oct 5, 2016

$x = \frac{1}{2} \left(\frac{2}{3} \pm \frac{\sqrt{14}}{3}\right)$

#### Explanation:

Making $y = \sin \left(x\right)$ and $\alpha = \frac{2}{3}$

$y + \sqrt{1 - {y}^{2}} = \alpha$
$1 - {y}^{2} = {\left(\alpha - y\right)}^{2}$
$1 - {y}^{2} = {\alpha}^{2} - 2 \alpha y + {y}^{2}$
$2 {y}^{2} - 2 \alpha y + {\alpha}^{2} - 1 = 0$ solving for $y$

$y = \frac{1}{2} \left(\alpha \pm \sqrt{2 - {\alpha}^{2}}\right)$

$\sin \left(x\right) = \frac{1}{2} \left(\frac{2}{3} \pm \frac{\sqrt{14}}{3}\right)$

Oct 5, 2016

Considering the given condition is

$\cos x + \sin x = \frac{2}{3.} \ldots . . \left(1\right)$

Now

${\left(\cos x - \sin x\right)}^{2} + {\left(\cos x + \sin x\right)}^{2} = 2 \left({\cos}^{2} x + {\sin}^{2} x\right) = 2$

$\implies {\left(\cos x - \sin x\right)}^{2} + {\left(\frac{2}{3}\right)}^{2} = 2$

$\implies {\left(\cos x - \sin x\right)}^{2} + {\left(\frac{2}{3}\right)}^{2} = 2$

$\implies {\left(\cos x - \sin x\right)}^{2} = 2 - \frac{4}{9} = \frac{14}{9}$

$\implies \left(\cos x - \sin x\right) = \pm \frac{\sqrt{14}}{3.} \ldots . \left(2\right)$

Adding (1) and (2) we get

$2 \cos x = \frac{2}{3} \pm \frac{\sqrt{14}}{3}$

$\implies \cos x = \frac{1}{2} \left(\frac{2}{3} \pm \frac{\sqrt{14}}{3}\right)$