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

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

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2(sin(x))^2-5cos(x)+1=0

2 (sin (x))^{2} - 5 cos (x) + 1 = 0

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

Solution steps

2 (sin (x))^{2} - 5 cos (x) + 1 = 0

Add

5 cos (x)

to both sides

2 sin^{2} (x) + 1 = 5 cos (x)

Square both sides

(2 sin^{2} (x) + 1)^{2} = (5 cos (x))^{2}

Subtract

(5 cos (x))^{2}

from both sides

(2 sin^{2} (x) + 1)^{2} - 25 cos^{2} (x) = 0

Factor

(2 sin^{2} (x) + 1)^{2} - 25 cos^{2} (x) : (2 sin^{2} (x) + 1 + 5 cos (x)) (2 sin^{2} (x) + 1 - 5 cos (x))

(2 sin^{2} (x) + 1 + 5 cos (x)) (2 sin^{2} (x) + 1 - 5 cos (x)) = 0

Solving each part separately

2 sin^{2} (x) + 1 + 5 cos (x) = 0 or 2 sin^{2} (x) + 1 - 5 cos (x) = 0

2 sin^{2} (x) + 1 + 5 cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

2 sin^{2} (x) + 1 - 5 cos (x) = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Verify solutions by plugging them into the original equation

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

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

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

9 cosh (x) - 5 sinh (x) = 15

3 sin (x) = 3 cos (x)

- 4 sin^{2} (θ) - 7 sin (θ) + 4 = 0

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