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

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

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2+cos(2x)=-3cos(x)

2 + cos (2 x) = - 3 cos (x)

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

Solution steps

2 + cos (2 x) = - 3 cos (x)

Subtract

- 3 cos (x)

from both sides

2 + cos (2 x) + 3 cos (x) = 0

Rewrite using trig identities

1 + 2 cos^{2} (x) + 3 cos (x) = 0

Solve by substitution

cos (x) = - \frac{1}{2}, cos (x) = - 1

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

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

tan (\frac{3 x}{2} + \frac{π}{2}) = 1

2 tan^{2} (x) + sec^{2} (x) = 2

3 sin (x) + 4 cos (x) = 5

4 cos (2 t) + 1 = 3, 0 \leq t \leq \frac{π}{2}

4 cos^{3} (x) = 3 cos (x)

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