Trig Function

**dwat** · June 5th 2009, 09:44 PM

How to solve this trig function $2cos(3x)=1$ for all the values of x: $\frac{-\pi}{6}\leq{x}\leq\frac{5\pi}{6}$

can someone please demonstrate how would I solve such problems step by step? I'm still trying to get my head around these trig functions.

**Random Variable** · June 5th 2009, 10:27 PM

$2cos(3x)=1$

$cos(3x)= \frac{1}{2}$

$cos^{-1}(cos(3x)) = cos^{-1}(\frac {1}{2})$

$3x = cos^{-1}( \frac {1}{2})$

You want to find the angle (or angles) in the range given whose cosine is 1/2. There are two angles: $\frac{\pi}{3}$ and $-\frac{\pi}{3}$

so $3x = \pm \frac{\pi}{3}$

or $x = \pm \frac{\pi}{9}$

**jamtheman** · June 5th 2009, 10:37 PM

Dont forget that if possible the angle may be +or- 2*pi within the domain

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