# Solving Trig Equations

• Mar 20th 2011, 03:42 PM
funkyfiasco
Solving Trig Equations
4cos2x = - √8

cos(3x + 180º) = √3/2

cos2x-sin2xsecx = 1

• Mar 20th 2011, 03:47 PM
pickslides
What have you tried with these?
• Mar 20th 2011, 03:49 PM
funkyfiasco
Well we're either factoring or using the trig identities (along with addition/half/double angle formulas)
• Mar 20th 2011, 04:30 PM
pickslides
Some identities could be helpful here, but maybe you don't need them at all.

Take #1 for example

$\displaystyle 4\cos 2x = -\sqrt{8}$

$\displaystyle \cos 2x = \frac{-\sqrt{8}}{4}$

$\displaystyle \cos 2x = \frac{-2\sqrt{2}}{4}$

$\displaystyle \cos 2x = \frac{-\sqrt{2}}{2}$

$\displaystyle \cos 2x = \frac{-1}{\sqrt{2}}$

$\displaystyle 2x = \frac{3\pi}{4}, \frac{5\pi}{4}$

$\displaystyle x = \frac{3\pi}{8}, \frac{5\pi}{8}$
• Mar 20th 2011, 04:31 PM
topsquark
Quote:

Originally Posted by funkyfiasco
Well we're either factoring or using the trig identities (along with addition/half/double angle formulas)

Okay, so for example what have you done with this one?
$4~cos(2x) = - \sqrt{8}$

-Dan

Edit: Slow on the guns again!
• Mar 20th 2011, 04:43 PM
funkyfiasco
for 4cos2x = -√8 I got 67.5 degrees and 112.5 degrees.

I'm still trying to find cos(3x + 180 deg) = √3/2 and the other one.

• Mar 20th 2011, 04:53 PM
e^(i*pi)
Quote:

Originally Posted by funkyfiasco

$\cos(3x + 180^{o}) = \dfrac{\sqrt{3}}{2}$
.

Use the addition formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$

$\cos(3x+180^o) = \cos(3x) \cos(180^o) - \sin(3x) \sin(180^o)$

180 is on the unit circle so you can simplify
• Mar 20th 2011, 06:46 PM
Prove It
Or you can use the fact that $\displaystyle \cos{(\theta + 180^{\circ})} = -\cos{\theta}$...