# Trigonometric equation

• Mar 5th 2011, 08:03 PM
isuckatallmath
Trigonometric equation
http://littleimg.com/files/48366_lpkmm/Eqn1.gif

Work shown so far:

http://littleimg.com/files/48367_wh0fi/Eqn2.gif

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?
• Mar 5th 2011, 08:22 PM
Prove It
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.
• Mar 5th 2011, 08:55 PM
isuckatallmath
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).
• Mar 5th 2011, 08:59 PM
Prove It
I believe you mean $\displaystyle \displaystyle \sin{x} = \frac{-1 \pm \sqrt{1 - 4(-2)(1)}}{2(-2)} = -\frac{1}{2}$ or $\displaystyle \displaystyle 1$...
• Mar 5th 2011, 09:06 PM
isuckatallmath
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?
• Mar 5th 2011, 11:27 PM
TheCoffeeMachine
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 $\displaystyle 2x = 2n\pi-x-\frac{\pi}{2} \Rightarrow x = \frac{2}{3}n\pi-\frac{\pi}{6}.$