4cos2x = - √8

cos(3x + 180º) = √3/2

cos2x-sin2xsecx = 1

Thanks in advance.

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- Mar 20th 2011, 03:42 PMfunkyfiascoSolving Trig Equations
4cos2x = - √8

cos(3x + 180º) = √3/2

cos2x-sin2xsecx = 1

Thanks in advance. - Mar 20th 2011, 03:47 PMpickslides
What have you tried with these?

- Mar 20th 2011, 03:49 PMfunkyfiasco
Well we're either factoring or using the trig identities (along with addition/half/double angle formulas)

- Mar 20th 2011, 04:30 PMpickslides
Some identities could be helpful here, but maybe you don't need them at all.

Take #1 for example

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

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

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

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

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

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

$\displaystyle \displaystyle x = \frac{3\pi}{8}, \frac{5\pi}{8}$ - Mar 20th 2011, 04:31 PMtopsquark
- Mar 20th 2011, 04:43 PMfunkyfiasco
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.

Thanks for your help - Mar 20th 2011, 04:53 PMe^(i*pi)
- Mar 20th 2011, 06:46 PMProve It
Or you can use the fact that $\displaystyle \displaystyle \cos{(\theta + 180^{\circ})} = -\cos{\theta}$...