Solving Trigonometric equation

**Mike Clemmons** · February 8th 2010, 04:31 PM

Hi,
Looking for help with this problem---I'm stumped because I get different answers than the book.
Here's the problem:
(2sin^2)(2x) = 1

The book gets the following answers: pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8,11pi/8, 13pi/8, 15pi/8.

I keep getting pi/4, 3pi/4, 5pi/4, 7pi/4.

Appreciate an explanation of how they get the answers.
Mike Clemmons

**Soroban** · February 8th 2010, 10:12 PM

Hello, Mike!

You're probably making a small error.

I assume that answers are in the interval $[0,\:2\pi]$

I'll take baby steps . . .

$2\sin^2(2x) \:=\: 1$

Divide by 2: . $\sin^2(2x) \:=\:\frac{1}{2}$

Take square roots: . $\sin(2x) \;=\;\pm\frac{1}{\sqrt{2}}$

(Can you see that the angle is a 45° angle in each quadrant?)

We have: . $2x \;=\;\frac{\pi}{4} + \frac{\pi}{2}n$

Divide by 2: . $x \;=\;\frac{\pi}{8} + \frac{\pi}{4}n$

Let $n \:=\:0,1,2,3,\hdots 7$

. . $x \;=\;\frac{\pi}{8},\;\frac{3\pi}{8},\;\frac{5\pi}{ 8},\;\frac{7\pi}{8},\;\frac{9\pi}{8},\;\frac{11\pi }{8},\;\frac{13\pi}{8},\;\frac{15\pi}{8}$

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