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

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

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

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

x = \frac{5 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn

Solution steps

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

Let:

u = \frac{x}{2}

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

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

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

Combine all the solutions

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

Substitute back

u = \frac{x}{2}

\frac{x}{2} = \frac{5 π}{6} + πn : x = \frac{5 π}{3} + 2 πn

\frac{x}{2} = \frac{π}{6} + πn : x = \frac{π}{3} + 2 πn

x = \frac{5 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn

1 = cos (θ)

sec (4 θ) - 2 = 0

3 sin (θ) = 1 + cos (θ)

cos (θ) = \frac{6}{10}

cos (x) \cdot tan (x) - cos (x) = 0

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