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

The general solution for sin^2(x)-4cos(x)+4=0 is x=2pin

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

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

x = 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = - 5, cos (x) = 1

cos (x) = - 5 :

No Solution

cos (x) = 1 : x = 2 πn

Combine all the solutions

x = 2 πn

sec (θ) = sin^{2} (θ) sec (θ)

sin (2 θ) + sin (θ) = 0, 0 \leq θ \leq 2 π

1 + cos (x) = 2 - cos (x)

cos^{2} (x) - 3 cos (x) - 4 = 0

2 sin^{2} (x) = sin (x) + 3

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