Sin Equation - Possibly needs Sum/Difference Identities??

**blueskywit** · June 17th 2009, 07:33 PM

Stuck on the following - SOLVE:

$2*sin^2(x) = 1 + 2*sin(x)$ over the interval $[0,2Pi)$

Factored with the quadratic to get:

$sin(x) = (1+sqrt(3))/2$
and
$sin(x) = (1-sqrt(3))/2$

But now I'm stuck - I've tried to apply sum/difference rules since 1/2 and sqrt(3)/2 are common values but still no luck.

Any Ideas??
Whit

**yeongil** · June 17th 2009, 09:12 PM

Assuming that these are correct:
$sin(x) = (1+sqrt(3))/2$
and
$sin(x) = (1-sqrt(3))/2$

You can immediately reject the first one, because $\frac{1+\sqrt{3}}{2}$ is greater than 1.

If you're wondering whether solving $\sin(x) = \frac{1-\sqrt{3}}{2}$ is supposed to give you a "nice" angle, the answer is no. I get x = -21.47 degrees (approx). I used the calculator to get that answer.

01

**VonNemo19** · June 17th 2009, 09:22 PM

Originally Posted by blueskywit

Stuck on the following - SOLVE:

$2*sin^2(x) = 1 + 2*sin(x)$ over the interval $[0,2Pi)$

Factored with the quadratic to get:

$sin(x) = (1+sqrt(3))/2$
and
$sin(x) = (1-sqrt(3))/2$

But now I'm stuck - I've tried to apply sum/difference rules since 1/2 and sqrt(3)/2 are common values but still no luck.

Any Ideas??
Whit

You're not stuck!

$2sin^2x-2sinx-1=0$

I'm not really sure how you factored this one

I'd use this

$sinx=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)}=\frac{2\pm\sqrt{12}}{4}=\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}$

So youre good up to here

So maybe try this

$sin(\frac{1}{2}+\frac{\sqrt{3}}{2})=sin\frac{1}{2} cos\frac{\sqrt{3}}{2}+sin\frac{\sqrt{3}}{2}cos\fra c{1}{2}$

**yeongil** · June 17th 2009, 09:35 PM

Originally Posted by VonNemo19

So maybe try this

$sin(\frac{1}{2}+\frac{\sqrt{3}}{2})=sin\frac{1}{2} cos\frac{\sqrt{3}}{2}+sin\frac{\sqrt{3}}{2}cos\fra c{1}{2}$

No, you don't want to try that. You're confusing the angles with the ratios.

When you get to this:
$\sin x = \frac{1-\sqrt{3}}{2}$

you should take the inverse sine of both sides (just remember the domain/range restrictions):
$\begin{aligned}<br /> \sin^{-1}(\sin x) &= \sin^{-1}\left(\frac{1-\sqrt{3}}{2}\right) \\<br /> x &= \sin^{-1}\left(\frac{1-\sqrt{3}}{2}\right) \\<br /> x &\approx -21.47^{\circ}<br /> \end{aligned}$

01

**blueskywit** · June 18th 2009, 05:13 AM

So just to make sure then, this equation will have NO solution on the interval [0,2Pi) ?

**yeongil** · June 18th 2009, 05:30 AM

Originally Posted by blueskywit

So just to make sure then, this equation will have NO solution on the interval [0,2Pi) ?

Oh, oops, I forgot that the solution had to be in $[0, 2\pi)$ , and that I should be in radians.

Just take the answer and add 360 degrees:
$x \approx -21.47^{\circ} + 360^{\circ} \approx 338.53^{\circ} \approx 5.91 rad$

01

**VonNemo19** · June 18th 2009, 07:01 PM

Originally Posted by yeongil

No, you don't want to try that. You're confusing the angles with the ratios

01

You're absolutely right. I can't believe I did that.

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