# Thread: Solving Trig Equations

1. ## Solving Trig Equations

I'm about done all my homework for this unit, but I seem to be stuck on my last 2 questions....any help would really be appreciated.

Solve each equation:
a) 2cos^2x + 3cosx -2 = 0

and
b) sinx = cos2x

My lesson book is not very good on explaining the process of solving for x and provided very few examples....which is why I'm stuck. I just want to hand in my homework tonight and worry about understanding it when I can actually ask the teacher for help (which won't be until Monday)

Thank you for any help you can give me!

2. 1) Factorize the LHS of equation

$\displaystyle (2\cos x - 1) ( \cos x + 2) = 0$
$\displaystyle => 2\cos x - 1 = 0$ or $\displaystyle \cos x + 2 = 0$
=> $\displaystyle 2\cos x = 1$ or $\displaystyle \cos x = -2$

but $\displaystyle \cos x$cannot be -2.
so$\displaystyle 2\cos x = 1$
or $\displaystyle \cos x = 1/2$

or x = 60 degree Answer

b) $\displaystyle \sin x = \cos 2x$
=>$\displaystyle sinx = 1 - 2\sin^2 x$

Proceed in above method. Factorize and calculate.

3. does the answer end up being
sinx = -1 and sinx = 1/2?

4. Originally Posted by lil_cookie
does the answer end up being
sinx = -1 and sinx = 1/2?
Yes
So x = -90 degree or x = 30 degree.

Cheers.

5. Better to write the answer in general form so as to include all possible solutions

6. Got it, Thank you all!