Sin^4x -cos^4x= cos3x Could you solve this?

Question

Sin^4x -cos^4x= cos3x Could you solve this?

4 Answers

Impact of this question

5411 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-05T16:00:44

$x = \frac{π}{5}$ $x= pi/5$
$x = \frac{3 π}{5}$ $x = (3pi)/5$
$x = π$ $x= pi$

Explanation:

We have:

$({sin}^{2} x + {cos}^{2} x) ({sin}^{2} x - {cos}^{2} x) = cos (3 x)$ $(sin^2x+ cos^2x)(sin^2x- cos^2x) = cos(3x)$

$1 ({sin}^{2} x - {cos}^{2} x) = cos (3 x)$ $1(sin^2x - cos^2x)= cos(3x)$

$- cos (2 x) = cos (3 x)$ $-cos(2x) = cos(3x)$

$0 = cos (3 x) + cos (2 x)$ $0 = cos(3x) + cos(2x)$

$0 = cos (2 x) cos (x) - sin (2 x) sin x + cos (2 x)$ $0 = cos(2x)cos(x) - sin(2x)sinx + cos(2x)$

$0 = (2 {cos}^{2} x - 1) cos x - 2 sin x cos x sin x + 2 {cos}^{2} x - 1$ $0 = (2cos^2x -1)cosx- 2sinxcosxsinx + 2cos^2x- 1$

$0 = 2 {cos}^{3} x - cos x - 2 {sin}^{2} x cos x + 2 {cos}^{2} x - 1$ $0 = 2cos^3x - cosx - 2sin^2xcosx + 2cos^2x - 1$

$0 = 2 {cos}^{3} x - cos x - 2 (1 - {cos}^{2} x) cos x + 2 {cos}^{2} x - 1$ $0 = 2cos^3x- cosx - 2(1- cos^2x)cosx + 2cos^2x - 1$

$0 = 2 {cos}^{3} x - cos x - 2 (cos x - {cos}^{3} x) + 2 {cos}^{2} x - 1$ $0 = 2cos^3x- cosx - 2(cosx - cos^3x) + 2cos^2x- 1$

$0 = 2 {cos}^{3} x - cos x - 2 cos x + 2 {cos}^{3} x + 2 {cos}^{2} x - 1$ $0 = 2cos^3x- cosx- 2cosx+ 2cos^3x +2cos^2x- 1$

$0 = 4 {cos}^{3} x + 2 {cos}^{2} x - 3 cos x - 1$ $0 = 4cos^3x + 2cos^2x - 3cosx -1$

Let $u = cos x$ $u = cosx$ .

$0 = 4 u^{3} + 2 u^{2} - 3 u - 1$ $0 = 4u^3 + 2u^2 - 3u - 1$

We see that $u = - 1$ $u = -1$ is a factor. Using synthetic division we get

$0 = (x + 1) (4 x^{2} - 2 x - 1)$ $0 = (x + 1)(4x^2 - 2x - 1)$

The equation $4 x^{2} - 2 x - 1 = 0$ $4x^2 - 2x - 1= 0$ may be solved using the quadratic formula.

$x = \frac{2 \pm \sqrt{2^{2} - 4 \cdot 4 \cdot - 1}}{2 \cdot 4}$ $x = (2 +- sqrt(2^2 - 4 * 4 * -1))/(2 * 4)$

$x = \frac{2 \pm \sqrt{20}}{8}$ $x = (2 +- sqrt(20))/8$

$x = \frac{1 \pm \sqrt{5}}{4}$ $x = (1 +- sqrt(5))/4$

$x \approx 0.809 or - 0.309$ $x ~~ 0.809 or -0.309$

Since $cos x = u$ $cosx = u$ , we get $x = \frac{π}{5}, \frac{3 π}{5}$ $x = pi/5, (3pi)/5$ and $π$ $pi$ .
Where $n$ $n$ is an integer.

The graph of $y_{1} = {sin}^{4} x - {cos}^{4} x$ $y_1 = sin^4x- cos^4x$ and $y_{2} = cos (3 x)$ $y_2 = cos(3x)$ confirms that the solutions are the intersection points.

enter image source here

Hopefully this helps!

Answer 2 · 2018-04-05T18:37:35

$x = (2 k + 1) π$ $x = (2k + 1)pi$
$x = \frac{(2 k - 1) π}{5}$ $x = ((2k - 1)pi)/5$

Explanation:

${sin}^{4} x - {cos}^{4} x = cos 3 x$ $sin^4x - cos^4 x = cos 3x$
$({sin}^{2} x + {cos}^{2} x) ({sin}^{2} x - {cos}^{2} x) = cos 3 x$ $(sin^2 x + cos^2 x)(sin^2 x - cos^2 x) = cos 3x$
$1 ({sin}^{2} x - {cos}^{2} x) = cos 3 x$ $1(sin^2 x - cos^2 x) = cos 3x$
$- cos 2 x = cos 3 x$ $-cos 2x = cos 3x$ , or
$cos 3 x = - cos 2 x = cos (2 x + π)$ $cos 3x = - cos 2x = cos (2x + pi)$
Unit circle, and property of cos, give -->
$3 x = \pm (2 x + π) + 2 k π$ $3x = +- (2x + pi) + 2kpi$
a. $3 x = 2 x + π + 2 k π$ $3x = 2x + pi + 2kpi$
$x = (2 k + 1) π$ $x = (2k + 1)pi$
If k = 0 --> $x = π$ $x = pi$
b. $3 x = - 2 x - π + 2 k π$ $3x = - 2x - pi + 2kpi$
$5 x = (2 k - 1) π$ $5x = (2k - 1)pi$ ,
$x = \frac{(2 k - 1) π}{5}$ $x = ((2k - 1)pi)/5$
If k = 1 --> $x = \frac{π}{5}$ $x = pi/5$ .
If k = 0 --> $x = - \frac{π}{5}$ $x = - pi/5$ , or $x = \frac{9 π}{5}$ $x = (9pi)/5$ (co-terminal)
If k = 2 --> $x = \frac{3 π}{5}$ $x = (3pi)/5$
In the closed interval [0, 2pi], the answers are:
$0, \frac{π}{5}, \frac{3 π}{5}, π, \frac{9 π}{5}$ $0, (pi)/5, (3pi)/5, pi, (9pi)/5$
Check by calculator.
$x = \frac{π}{5} = 36^{\circ}$ $x = pi/5 = 36^@$ --> ${sin}^{4} x = 0.119$ $sin^4 x = 0.119$ --> ${cos}^{4} x = - 0.428$ $cos^4 x = - 0.428$ --> cos 3x = - 309.
${sin}^{4} x - {cos}^{4} x = 0.119 - 0.428 = - 309$ $sin^4 x - cos^4 x = 0.119 - 0.428 = - 309$ . Proved
$x = \frac{9 π}{5}$ $x = (9pi)/5$ --> ${sin}^{4} x = 0.119$ $sin^4 x = 0.119$ --> ${cos}^{4} x = 0.428$ $cos^4 x = 0.428$ -->
${sin}^{4} x - {cos}^{4} x = - 0.309$ $sin^4 x - cos^4 x = - 0.309$
$cos 3 x = cos 972 = - 0.309$ $cos 3x = cos 972 = - 0.309$ . Proved

Answer 3 · 2018-04-13T04:25:50

$\to x = (2 n + 1) \frac{π}{5}, (2 n + 1) π$ $rarrx=(2n+1)pi/5, (2n+1)pi$ $n \to Z$ $nrarrZ$

Explanation:

$\to {sin}^{4} x - {cos}^{4} x = cos 3 x$ $rarrsin^4x-cos^4x=cos3x$

$\to ({sin}^{2} x + {cos}^{2} x) ({sin}^{2} x - {cos}^{2} x) = cos 3 x$ $rarr(sin^2x+cos^2x)(sin^2x-cos^2x)=cos3x$

$\to - cos 2 x = cos 3 x$ $rarr-cos2x=cos3x$

$\to cos 3 x + cos 2 x = 0$ $rarrcos3x+cos2x=0$

$\to 2 cos (\frac{3 x + 2 x}{2})) \cdot cos (\frac{3 x - 2 x}{2})) = 0$ $rarr2cos((3x+2x)/2))*cos((3x-2x)/2))=0$

$\to cos (\frac{5 x}{2}) \cdot cos (\frac{x}{2}) = 0$ $rarrcos((5x)/2)*cos(x/2)=0$

Either $cos (\frac{5 x}{2}) = 0$ $cos((5x)/2)=0$

$\to \frac{5 x}{2} = (2 n + 1) \frac{π}{2}$ $rarr(5x)/2=(2n+1)pi/2$

$\to x = (2 n + 1) \frac{π}{5}$ $rarrx=(2n+1)pi/5$ $n \to Z$ $nrarrZ$

$\to cos (\frac{x}{2}) = 0$ $rarrcos(x/2)=0$

$\to \frac{x}{2} = (2 n + 1) \frac{π}{2}$ $rarrx/2=(2n+1)pi/2$

$\to x = (2 n + 1) π$ $rarrx=(2n+1)pi$ $n \to$ $nrarr$

Answer 4 · 2018-04-29T14:42:15

The general solution doesn't require the triple angle formula, and is

$x = 180^{\circ} + 360^{\circ} k$ $x=180 ^circ + 360^circ k$ or $x = 36^{\circ} + 72^{\circ} k$ $x = 36^circ + 72^circ k$

for integer $k$ $k$ .

Explanation:

I don't like reading other people's answers before I solve a question myself. But a featured answer for this one popped up. During my quick glance I couln't help notice it looked pretty complicated for what looks to me like a relatively easy question. I'll give it a shot.

${sin}^{4} x - {cos}^{4} x = cos 3 x$ $sin ^4 x - cos ^4 x = cos 3x$

$({sin}^{2} x + {cos}^{2} x) ({sin}^{2} x - {cos}^{2} x) = cos 3 x$ $(sin ^2 x + cos ^2 x)(sin ^2 x - cos ^2 x ) = cos 3x$

$- cos 2 x = cos 3 x$ $-cos 2x = cos 3x$

$cos (180^{\circ} - 2 x) = cos 3 x$ $cos (180^circ - 2x) = cos 3x$

I've been on Socratic for a couple of weeks, and this is emerging as my theme: The general solution to $cos x = cos a$ $cos x = cos a$ is $x = \pm a + 360 ^circ k quad$ for integer $k.$

$180^circ - 2x = \pm 3x + 360^circ k$

$-2x \pm 3x = -180^circ + 360^circ k$

We take the signs separately. Plus first:

$x = -180^circ + 360^circ k = 180 ^circ + 360^circ k$

Minus next.

$-5x = -180^circ + 360^circ k$

$x = 36^circ + 72^circ k$

If you read these closely you might think I'm making a mistake with the way I manipulate $k$ . But since $k$ ranges over all the integers, substitutions like $k to -k$ and $k to k+1$ are allowed and I slip those in to keep the signs $+$ when they can be.

Check:

Let's pick a couple to check. I'm geeky enough to know $cos 36^circ$ is half the Golden Ratio, but I'm not going to work these out exactly, just pop them into Wolfram Alpha to make sure.

$x = 36^circ + 72^circ = 108^circ$

$sin ^4 108 - cos ^4 108 - cos (3 * 108) = 0 quad sqrt$

$x = 180 - 2 (360) = -540$

$sin ^4 (-540) - cos^4(-540) - cos(3 * -540) = 0 quad sqrt$

Science

Math

Humanities

... and beyond

Sin^4x -cos^4x= cos3x Could you solve this?

4 Answers

Explanation:

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question