# Solving Trigonometric Equations (Advanced Functions)

• November 30th 2008, 02:14 PM
mehek
Determine the solution for this equation on the interval [ 0 is less or equal to x which is less than or equal to 2pi ] :

sin2x = 1/√2

How do you solve this question? We are supposed to solve for x and then graph it where sin will be positive or negative... I understand how to do equations with the simple trig functions, but not with the compound angle formulae...

If anyone could help me out with this question, that would be wonderful!! :)

EDIT: I just realized that we can't double post our questions... If any mod wants to delete this thread that would be great (since I already posted this in the General High School Homework Help topic)
• November 30th 2008, 02:19 PM
TheEmptySet
Quote:

Originally Posted by mehek
Determine the solution for this equation on the interval [ 0 is less or equal to x which is less than or equal to 2pi ] :

sin2x = 1/√2

How do you solve this question? We are supposed to solve for x and then graph it where sin will be positive or negative... I understand how to do equations with the simple trig functions, but not with the compound angle formulae...

If anyone could help me out with this question, that would be wonderful!! :)

Take the inverse sin of both sides

$\sin(2x)= \frac{1}{\sqrt{2}} \implies \sin^{-1}\left( \sin(2x)\right)=\sin^{-1}\left( \frac{1}{\sqrt{2}}\right)$

this gives us two equations

$2x=\frac{\pi}{4}$ or $2x=\frac{3\pi}{4}$

I hope this helps. good luck
• November 30th 2008, 03:15 PM
skeeter
$0 \leq x \leq 2\pi$

$0 \leq 2x \leq 4\pi$

$2x = \frac{\pi}{4}$

$2x = \frac{3\pi}{4}$

$2x = \frac{9\pi}{4}$

$2x = \frac{11\pi}{4}$