1. ## solving trig equations

I have the question and answer to this question but i do not understand at all how the answer is concluded

Q: in the range 0<=x<=2Pi solve Sin2x=-1
A:3Pi/8 and 7Pi/8

could somebody maybe briefly explain how you work this out

thanks

2. For angles ranging from 0 to 360,
Sin(angle) may be viewed as the y co-ordinate of a point on a unit circle,

If 2x is the angle,
where is the y co-ordinate -1 ?

answer: at 270 degrees, or $\displaystyle \frac{3}{4}2\pi$

Sin(270 degrees)= -1.

So 2x is 270 degrees or 270+n(360) degrees, n=0,1,2,3........

x is $\displaystyle \frac{3}{4}\pi$

The second result is obtained for 2x=270+360 degrees,
as x<360 degrees.

3. Hello, wahhdoe!

Solve: .$\displaystyle \sin2x \:=\:\text{-}1\;\;\text{ in the range }\,0 \leq x \leq 2\pi$

Answers: .$\displaystyle \frac{3\pi}{8},\;\frac{7\pi}{8}$ . These are wrong! **

We are expect to know that: .$\displaystyle \sin270^o \:=\:\sin\frac{3\pi}{2} \:=\:-1$

We have: .$\displaystyle \sin2x \:=\:-1$

. . Then: .$\displaystyle 2x \;=\;\frac{3\pi}{2},\;\frac{7\pi}{2},\;\frac{11\pi }{2},\;\frac{15\pi}{2},\;\hdots$

. . Hence: .$\displaystyle x \;=\;\frac{3\pi}{4},\;\frac{7\pi}{4},\;\frac{11\pi }{4},\;\frac{15\pi}{4},\;\hdots$

The roots in the given range are: .$\displaystyle x \;=\;\frac{3\pi}{4},\;\frac{7\pi}{4}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Maybe the problem was: .$\displaystyle \sin{\color{red}4}x \:=\:-1$

Even then, their answer is wrong.
. . There are more answers in that range.

If the problem is: .$\displaystyle \sin4x \:=\:-1$

. . then we have: .$\displaystyle 4x \;=\;\frac{3\pi}{2},\;\frac{7\pi}{2},\;\frac{11\pi }{2},\;\frac{15\pi}{2},\;\hdots$

Therefore: .$\displaystyle x \;=\;\frac{3\pi}{8},\;\frac{7\pi}{8},\;\frac{11\pi }{8},\;\frac{15\pi}{8}$