Trigonometric equation

**isuckatallmath** · March 5th 2011, 08:03 PM

Work shown so far:

This is where I'm stuck at. I know that I somehow have to get two terms to equal to zero, but I don't know how. The +1 at the end of the problem is throwing me for a loop. Am I even doing it correct?

**Prove It** · March 5th 2011, 08:22 PM

Correct up to $\displaystyle 2 - 2\sin^2{x} + \sin{x} - 1 = 0$ .

From here, you have forgotten that after that it is an EQUATION, and also there should be a negative on the $\displaystyle 2\sin^2{x}$ .

Solve it as you would a quadratic equation.

**isuckatallmath** · March 5th 2011, 08:55 PM

Is this correct?

Also if it is, how do you know when to add the +n(2pi) to each of the solutions? It asked to solve the equation on interval [0,2pi).

**Prove It** · March 5th 2011, 08:59 PM

I believe you mean $\displaystyle \sin{x} = \frac{-1 \pm \sqrt{1 - 4(-2)(1)}}{2(-2)} = -\frac{1}{2}$ or $\displaystyle 1$ ...

**isuckatallmath** · March 5th 2011, 09:06 PM

yessir that's what I mean hehe. Still stuck in algebra mode. apologies. Thanks again for ur help Prove It. Do you happen to know the answer to the question about the add on?

**TheCoffeeMachine** · March 5th 2011, 11:27 PM

Originally Posted by isuckatallmath

By the identity $1+\cos{2x} = 2\cos^2{x}$ , we've $2\cos^2{x}+\sin{x}-1 = \cos{2x}+\sin{x}$ .
So for your equation we have $\cos{2x}+\sin{x} = 0 \Rightarrow \cos{2x} = -\sin{x} = \cos\left(x+\frac{\pi}{2}\right)$ .
If $\cos{\theta} = \cos{\alpha}$ , then $\theta = 2n\pi\pm \alpha$ , thus for the equation $\cos{2x} = \cos\left(x+\frac{\pi}{2}\right)$ we've:
$2x = 2n\pi+x+\frac{\pi}{2} \Rightarrow x = 2n\pi+\frac{\pi}{2}$ or we've $2x = 2n\pi-x-\frac{\pi}{2} \Rightarrow x = \frac{2}{3}n\pi-\frac{\pi}{6}.$

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