# solving trig equations

• Jan 18th 2010, 05:12 AM
wahhdoe
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
• Jan 18th 2010, 05:27 AM
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 $\frac{3}{4}2\pi$

Sin(270 degrees)= -1.

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

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

The second result is obtained for 2x=270+360 degrees,
as x<360 degrees.
• Jan 18th 2010, 06:18 AM
Soroban
Hello, wahhdoe!

Quote:

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

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

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

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

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

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

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

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

**

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

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

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

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

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