What is the general solution for 2cos^2(x)+5sin(x)-4=0 ?

The general solution for 2cos^2(x)+5sin(x)-4=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2cos^2(x)+5sin(x)-4=0

2 cos^{2} (x) + 5 sin (x) - 4 = 0

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

2 cos^{2} (x) + 5 sin (x) - 4 = 0

Rewrite using trig identities

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

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = 2

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = 2 :

No Solution

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

cos^{2} (\frac{x}{2}) - sin^{2} (\frac{x}{2}) = sin (x)

tan (x) + sec (x) = - 1

2 sin^{2} (θ) - cos (θ) - 1 = 0

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

What is the general solution for 2cos^2(x)+5sin(x)-4=0 ?
The general solution for 2cos^2(x)+5sin(x)-4=0 is x= pi/6+2pin,x=(5pi)/6+2pin