# Trigonometry - Checking my work

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Find the sum of the three smallest positive values of theta such that 4 cos^2(2theta-pi) =3. (Give your answer in radians.)

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 - ,

Hmmm.
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 - ,

Oh thanks that helped a lot :)

• Trigonometry - Checking my work - ,

if sinA= 1/3, then find sin (A+pi/6), cos(A-pi/3), tan(A-pi4)