
Trigonometric equation
http://littleimg.com/files/48366_lpkmm/Eqn1.gif
Work shown so far:
http://littleimg.com/files/48367_wh0fi/Eqn2.gif
(Headbang)
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?

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.

Is this correct?
http://littleimg.com/files/48368_u2q7p/Eqn4.gif
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).

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

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?

Quote:
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 $\displaystyle 2x = 2n\pix\frac{\pi}{2} \Rightarrow x = \frac{2}{3}n\pi\frac{\pi}{6}.$