1. ## Trigonometric equation

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?

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

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

Solve it as you would a quadratic equation.

3. 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).

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

5. 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?

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