# Trig

• Oct 24th 2013, 11:59 PM
bookish12
Trig

The solution set is _______ For example if there are 3 solutions, then write {0, pi/3, 3pi/4}

• Oct 25th 2013, 12:46 AM
MathoMan
Re: Trig
Quote:

Originally Posted by bookish12

The solution set is _______ For example if there are 3 solutions, then write {0, pi/3, 3pi/4}

$\displaystyle 2\sin(2x)=1\Rightarrow \sin(2x)=\frac{1}{2}\Rightarrow 2x=\frac{\pi}{6}+2k\pi \text{ or } 2x=\frac{5\pi}{6}+2k\pi\Rightarrow\\ \Rightarrow x=\frac{\pi}{12}+k\pi \text{ or } x=\frac{5\pi}{12}+k\pi, k\in\mathbb{Z}.$

Since $\displaystyle 0\leq x\leq 2\pi \Leftrightarrow x\in[0,2\pi]=\left$0, \frac{24\pi}{12} \right$$ you just have to check for what values of $\displaystyle k\in\mathbb{Z}$ solutions given above fall into the interval of interest.
• Oct 25th 2013, 05:19 AM
Soroban
Re: Trig
Hello, bookish12!

I assume that the second "2" is an exponent.

Quote:

$\displaystyle \text{Solve: } 2\sin^2(x) \:=\: 1,\;\;0 \le x < 2\pi$

$\displaystyle \begin{array}{c}2\sin^2(x) \:=\:1 \\ \\ \sin^2(x) \:=\:\frac{1}{2} \\ \\ \sin(x) \:=\:\pm\dfrac{1}{\sqrt{2}} \\ \\ x \;=\;\dfrac{\pi}{4},\,\dfrac{3\pi}{4},\,\dfrac{ 5\pi}{4},\,\dfrac{7\pi}{4} \end{array}$
• Oct 26th 2013, 01:50 PM
bookish12
Re: Trig
Thank you!!