Various trigonometric equations

**littlelisa** · May 19th 2008, 05:18 PM

Hello, I have this worksheet and there are a few problems I need clarification on. I've done most of them but my answers all turn out weird.

If cos(x) = -7/9, tan(x) < 0, find cos2(x)

Find all the solutions of sin5x + sinx = sin3x in the interval [0, 2pi)

Rewrite sin3xcos4y as a sum

Find all the solutions of 3cos(x/4) in the interval of [0, 2pi)

Thanks

**Soroban** · May 19th 2008, 05:52 PM

Hello, littlelisa!

If $\cos x = \text{-}\frac{7}{9},\;\;\tan x < 0$ , find $\cos2x$

Cosine is negative in Quadrants 2 and 3.
Tangent is negative in Quadrants 2 and 4.
. . Hence, $x$ is in Quadrant 2.

Identity: . $\cos2\theta \:=\:2\cos^2\!\theta -1$

So we have: . $\cos2x \;=\;2\left(\text{-}\frac{7}{9}\right)^2 - 1 \;=\;\frac{17}{81}$

Find all the solutions of $\sin5x + \sin x \:=\: \sin3x$ in the interval $[0, 2\pi)$

Sum-to-product identity: . $\sin A + \sin B \;=\;2\sin\left(\frac{A+B}{2}\right)\cos\left(\fra c{A-B}{2}\right)$

So: . $\sin5x+\sin x \:=\:2\sin\left(\frac{5x+x}{2}\right)\cos\left(\fr ac{5x-x}{2}\right) \;=\;2\sin3x\cos2x$

The equation becomes: . $2\sin3x\cos2x\:=\:\sin3x \quad\Rightarrow\quad 2\sin3x\cos2x - \sin3x \:=\:0$

. . Factor: . $\sin3x(2\cos2x - 1) \:=\:0$

We have two equations to solve . . .

$\sin3x \:=\:0\quad\Rightarrow\quad 3x \:=\:0, \pi, 2\pi, 3\pi, 4\pi, 5\pi$
. . Therefore: . $\boxed{x \:=\:0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}}$

$2\cos2x-1\:=\:0\quad\Rightarrow\quad \cos2x \:=\:\frac{1}{2}\quad\Rightarrow\quad 2x \:=\:\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3},\ frac{11\pi}{3}$
. . Therefore: . $\boxed{x \;=\;\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}}$

Rewrite $\sin3x\cos4y$ as a sum.

Product-to-sum identity: . $\sin A\cos B \;=\;\frac{1}{2}\bigg[\sin(A-B) + \sin(A + B)\bigg]$

Therefore: . $\sin3x\cos4y \;=\;\frac{1}{2}\sin(3x-4y) + \frac{1}{2}\sin(3x+4y)$

Find all the solutions of $3\cos\left(\frac{x}{4}\right)$ in the interval $[0, 2\pi)$

You left out part of the equation . . .

**littlelisa** · May 19th 2008, 06:50 PM

Thanks and sorry about the last equation it equals 0.

3cos(x/4)=0

Thanks again I really appreciate it!!!

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