Trig Problem: Finding Principle Values of x

[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
?

 

[pizzasky]

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.

In your working, you wrote:
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.

 

[HallsofIvy]

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

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.

$$<b> <u>sinx=1</u></b>$$

[tex] Okay, this at least makes sense but I don't see where you got it from<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> (2sin/2)= <br /> <b> <u>sin=½</u></b> </div> </div> </blockquote> sine is a <b>function</b>! It makes no sense to say "2sin/2" or <br /> "sin= 1/2".<br /> sin(x)= 1/2. But you still haven't answered the question. What is x?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 1b. 0°≤x≥180°<br /> <br /> sin1/2=±√1-cosx/2<br /> 2sin²x-1=0<br /> 2sin²x=1<br /> 2sin(sinx)=1<br /> 2sin/2=1/2<br /> sin=1/2 </div> </div> </blockquote>I have absolutely no idea what you are doing here!<br /> 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). <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <u><b>sin½= 30°,90°</b></u> </div> </div> </blockquote>Once again, makes no sense! According to my calculator, sin(1/2 (radian))= 0.47942553860420300027328793521557. If you mean arcsin(1/2) or <br /> sin^-1(1/2), say so! In any case, sin(90°)= 1, not 1/2.<br /> 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°.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 1c. all values of x<br /> ? </div> </div> </blockquote> 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.[/tex]

 

[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.