Solving Trigonometric Equations (Advanced Functions)

**mehek** · November 30th 2008, 02:14 PM

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)

**TheEmptySet** · November 30th 2008, 02:19 PM

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