What is the general solution for cos^2(x)-3/2 cos(2x)=0 ?

The general solution for cos^2(x)-3/2 cos(2x)=0 is x=(5pi)/6+pin,x= pi/6+pin

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cos^2(x)-3/2 cos(2x)=0

cos^{2} (x) - \frac{3}{2} cos (2 x) = 0

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

Solution steps

cos^{2} (x) - \frac{3}{2} cos (2 x) = 0

Simplify

cos^{2} (x) - \frac{3}{2} cos (2 x) : \frac{2 cos ^{2} ( x ) - 3 cos ( 2 x )}{2}

\frac{2 cos ^{2} ( x ) - 3 cos ( 2 x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 cos^{2} (x) - 3 cos (2 x) = 0

Rewrite using trig identities

- cos^{2} (x) + 3 sin^{2} (x) = 0

Factor

- cos^{2} (x) + 3 sin^{2} (x) : (3 sin (x) + cos (x)) (3 sin (x) - cos (x))

(3 sin (x) + cos (x)) (3 sin (x) - cos (x)) = 0

Solving each part separately

3 sin (x) + cos (x) = 0 or 3 sin (x) - cos (x) = 0

3 sin (x) + cos (x) = 0 : x = \frac{5 π}{6} + πn

3 sin (x) - cos (x) = 0 : x = \frac{π}{6} + πn

Combine all the solutions

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

- 12 - 3 tan (θ) = - 12 + 3

12 cosh (2 x) + 7 sinh (x) - 24 = 0

sec^{2} (θ) - 1 = 0

2 sin (θ) cos (θ) = - sin (θ)

6 cos (c) - 4 = cos (c) - 9

What is the general solution for cos^2(x)-3/2 cos(2x)=0 ?
The general solution for cos^2(x)-3/2 cos(2x)=0 is x=(5pi)/6+pin,x= pi/6+pin