What is the general solution for 2cos^2(θ)-7cos(θ)-4=0 ?

The general solution for 2cos^2(θ)-7cos(θ)-4=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2cos^2(θ)-7cos(θ)-4=0

2 cos^{2} (θ) - 7 cos (θ) - 4 = 0

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

Solution steps

2 cos^{2} (θ) - 7 cos (θ) - 4 = 0

Solve by substitution

cos (θ) = 4, cos (θ) = - \frac{1}{2}

cos (θ) = 4 :

No Solution

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

Combine all the solutions

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

sin (θ) = \frac{3}{10}, sin (θ + \frac{π}{6})

3 sin^{2} (x) = 8 cos (x)

7 cos (x) - 24 sin (x) = 10

4 cos (2 θ) + 1 = 2 cos (θ)

sin (2 x) = - 2 cos^{2} (x), 0 \leq x \leq 2 π

What is the general solution for 2cos^2(θ)-7cos(θ)-4=0 ?
The general solution for 2cos^2(θ)-7cos(θ)-4=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin