SOLUTION: Find all solutions in the interval [0, 2π). cosθcos2θ+sinθsin2θ= sqrt2/2

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Question 1030520: Find all solutions in the interval [0, 2π). cosθcos2θ+sinθsin2θ= sqrt2/2
Found 2 solutions by Cromlix, ikleyn:
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
cosθcos2θ + sinθsin2θ = √2/2
cosθ(2cos^2θ - 1) + sinθ(2sinθcosθ)= √2/2
2cos^3θ - cosθ + 2sin^2θcosθ = √2/2
2cos^3θ - cosθ + 2(1 - cos^2θ)cosθ = √2/2
2cos^3θ - cosθ + 2cosθ - 2cos^3θ = √2/2
Cancel 2cos^3θ's
-cosθ + 2cosθ = √2/2
cosθ = √2/2
θ = 0.8, 5.5 radians.
Hope this helps :-)

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find all solutions in the interval [0, 2π). cosθcos2θ+sinθsin2θ= sqrt2/2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Use the formula  cos%28alpha-beta%29 = cos%28alpha%29%2Acos%28beta%29+%2B+sin%28alpha%29%2Asin%28beta%29

(see, for example, the lesson Addition and subtraction formulas in this site).

It gives you 

cos%28theta%29%2Acos%282theta%29+%2B+sin%28theta%29%2Asin%282theta%29 = cos%28theta-2theta%29 = cos%28-theta%29.

So, your original equation becomes

cos%28-theta%29 = sqrt%282%29%2F2.   (1)

Since  cos%28-theta%29  is always equal to  cos%28theta%29, you can rewrite (1) in the form

cos%28theta%29 = sqrt%282%29%2F2.   (2)

The equation (2) has two and only two solutions in the given interval.
They are  theta = pi%2F4  and/or  theta = 7pi%2F4.