Wednesday

March 4, 2015

March 4, 2015

Posted by **Sam** on Wednesday, October 30, 2013 at 9:32am.

Hi guys, I've been struggling with this problem. Here's my thinking about the problem:

we can rewrite the equation as cos^2(2theta-pi) = 3/4

Then we sqrt both sides --> cos(2theta - pi) = sqrt3 / 2

we find what arccos sqrt3 / 2 ---> pi/6 radians or 45 degrees (we'll use radians)

thus cos (2theta - pi) = cos(pi/6)

so 2 theta - pi = pi/6

and 2 theta = 7pi/6 ---> 7pi/12

When we let arccos sqrt3/2 ---> 11pi/6 and 13pi/6 we get 17pi/12 and 19pi/12 (those are the smallest two values which are positive, meaning 11pi/6 and 13pi/6)

so in all we get 17pi/12 + 19pi/12 + 7pi/12 = 43pi/12

Is this right?

- Trigonometry - Checking my work -
**Steve**, Wednesday, October 30, 2013 at 11:06amHmmm.

4cos^2(2θ-pi) = 3

cos^2(2θ-pi) = 3/4

since cos(2θ-pi) - -cos2θ, this makes things simpler, so we have

cos^2 2θ = 3/4

cos2θ = ±√3/2

2θ = π/6,5π/6,7π/6,...

θ = π/12,5π/12,7π/12

Looks like the sum is 13π/12

- Trigonometry - Checking my work -
**Sam**, Wednesday, October 30, 2013 at 11:11amOh thanks that helped a lot :)

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