What is the general solution for 3+2cos(x)=4cos(x/2) ?

The general solution for 3+2cos(x)=4cos(x/2) is x=(2pi)/3+4pin,x=(10pi)/3+4pin

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3+2cos(x)=4cos(x/2)

3 + 2 cos (x) = 4 cos (\frac{x}{2})

x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

Solution steps

3 + 2 cos (x) = 4 cos (\frac{x}{2})

Subtract

4 cos (\frac{x}{2})

from both sides

3 + 2 cos (x) - 4 cos (\frac{x}{2}) = 0

Let:

u = \frac{x}{2}

3 + 2 cos (2 u) - 4 cos (u) = 0

Rewrite using trig identities

1 - 4 cos (u) + 4 cos^{2} (u) = 0

Solve by substitution

cos (u) = \frac{1}{2}

cos (u) = \frac{1}{2} : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

Substitute back

u = \frac{x}{2}

\frac{x}{2} = \frac{π}{3} + 2 πn : x = \frac{2 π}{3} + 4 πn

\frac{x}{2} = \frac{5 π}{3} + 2 πn : x = \frac{10 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

- 3 sin (x) = cos (x)

2 csc^{2} (x) = 5 - 5 cot (x)

sin (x + 0.5) = 0.92, 0 \leq x \leq 2 π

cos (4 x) = 0.8 - 0.6

5 cos^{2} (x) = 4 - 3 sin^{2} (x)

What is the general solution for 3+2cos(x)=4cos(x/2) ?
The general solution for 3+2cos(x)=4cos(x/2) is x=(2pi)/3+4pin,x=(10pi)/3+4pin