What is the general solution for 2cos^2(θ)+3cos(θ)+4=3 ?

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

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2cos^2(θ)+3cos(θ)+4=3

2 cos^{2} (θ) + 3 cos (θ) + 4 = 3

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

Solution steps

2 cos^{2} (θ) + 3 cos (θ) + 4 = 3

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 sin (3 θ) cos (2 θ) - cos (2 θ) = 0

2 sec (x) + 1 = sec (x) + 3

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

- 7 sin (x) = 0

2 sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

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