What is the general solution for 2sin^2(θ)-5cos(θ)+1=0 ?

The general solution for 2sin^2(θ)-5cos(θ)+1=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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2sin^2(θ)-5cos(θ)+1=0

2 sin^{2} (θ) - 5 cos (θ) + 1 = 0

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

Solution steps

2 sin^{2} (θ) - 5 cos (θ) + 1 = 0

Rewrite using trig identities

3 - 2 cos^{2} (θ) - 5 cos (θ) = 0

Solve by substitution

cos (θ) = - 3, cos (θ) = \frac{1}{2}

cos (θ) = - 3 :

No Solution

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

Combine all the solutions

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

sec^{2} (x) cos (x) = cos (x), 0^{\circ} < x < 36 0^{\circ}

5 sin (x) - 4 = 2 cos (2 x)

csc (x) + cot (x) = 0

cos^{2} (θ) - sin^{2} (θ) = 0

7 = 4 sin (\frac{π}{2} (x - 2)) + 5

What is the general solution for 2sin^2(θ)-5cos(θ)+1=0 ?
The general solution for 2sin^2(θ)-5cos(θ)+1=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin