# Trig principle values problem

1. May 14, 2006

### WillyTech

I have a trig problem that i need help with. I have done a) & b) but not c) because i don't get it.

Solve the following for:

a) principle values of x
b) 0°≤x≥180°
c) all values of x

1. 2sin²x-1=0

my work:

1a. principle values of x

sinx(2sin)-1 = 0

sinx(2sin)=1

sinx=1

(2sin/2)=
sin=½

1b. 0°≤x≥180°

sin1/2=±√1-cosx/2
2sin²x-1=0
2sin²x=1
2sin(sinx)=1
2sin/2=1/2
sin=1/2

sin½= 30°,90°

1c. all values of x
?????

Last edited: May 14, 2006
2. May 14, 2006

Your initial working should be the same for all 3 parts, it's just the difference in the range of x for each part that results in you having different answers.

2sin²x-1=0
sinx(2sinx)=1 **
sinx=1 or 2sinx=1
sinx=1 or sinx=0.5

This does not seem right! You can only use this method if the right hand side of equation ** equals zero. Can you think of another way to do this? Note that there is only 1 sinx term in the original equation.

Also, do you know what the principal region for sinx is? For part b, I believe the range should be 0°≤x≤180°. And finally, for part c, you need to find a general solution which can "generate" all the possible values of x.

Please also take note of your expressions. For your working in part b, expressions like "2sin(sinx)=1" and "sin½= 30°,90°" don't seem very right.

Last edited: May 14, 2006
3. May 15, 2006

### HallsofIvy

Staff Emeritus
Please, please, please, learn to write exactly what you mean! I have no idea what "sinx(2sin)" could mean! If you don't want to use Tex or html codes, you could just say sin^2(x)= 1.

Okay, this at least makes sense but I don't see where you got it from

sine is a function! It makes no sense to say "2sin/2" or
"sin= 1/2".
sin(x)= 1/2. But you still haven't answered the question. What is x?

I have absolutely no idea what you are doing here!
The "principal" value is the single value of x between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ that gives sin(x)= 1/2. Since that is positive, yes, x must be positive. You still haven't said what that is. You can't do (b) until you have done (a).

Once again, makes no sense! According to my calculator, sin(1/2 (radian))= 0.47942553860420300027328793521557. If you mean arcsin(1/2) or
sin-1(1/2), say so! In any case, sin(90°)= 1, not 1/2.
However, you want 2sin2(x)- 1= 0 or $sin(x)= \frac{1}{\sqrt{2}}$. My calculator gives, as principal value (part (a) again!), x= 45°. You should know that sin(180°- x)= sin(x) so the other value, in $0\le x \le 180$ is 180- 45= 135°.

sine is periodic with period 360°. Take the two values you got for x between 0 and 180° (the only two solution between 0 and 360°) and add 360n to them.

4. May 15, 2006

### WillyTech

OOPS!! i guest i made a lot of mistakes.
ill go back over my work and reread the chapter, and thanks for checking my answers.

oh yeah, and it was 0≤x≤180................... i am really sorry i mistyped it.