What is the general solution for 5cos(2θ)=-15cos(θ)-10 ?

The general solution for 5cos(2θ)=-15cos(θ)-10 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin

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5cos(2θ)=-15cos(θ)-10

5 cos (2 θ) = - 15 cos (θ) - 10

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

Solution steps

5 cos (2 θ) = - 15 cos (θ) - 10

Subtract

- 15 cos (θ) - 10

from both sides

5 cos (2 θ) + 15 cos (θ) + 10 = 0

Rewrite using trig identities

5 + 10 cos^{2} (θ) + 15 cos (θ) = 0

Solve by substitution

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

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

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

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

2 cos (x) - 5 = 3 sec (x)

sin (3 θ + \frac{π}{18}) = 1, 0 \leq θ < 2 π

1 = tan (c)

sin (a) = \frac{3}{2}

tan^{2} (x) = 4

What is the general solution for 5cos(2θ)=-15cos(θ)-10 ?
The general solution for 5cos(2θ)=-15cos(θ)-10 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin