What is the general solution for sin^2(x)=2cos(x)+2,0<= x<= 2pi ?

The general solution for sin^2(x)=2cos(x)+2,0<= x<= 2pi is x=pi

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sin^2(x)=2cos(x)+2,0<= x<= 2pi

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

x = π

Solution steps

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

Subtract

2 cos (x) + 2

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = - 1

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

Combine all the solutions

x = π

tan^{4} (x) = tan^{4} (x) cos (x)

tan (x) sec (x) = 0

sin (x) + cos (x) = 1.2, sin (2 x)

2 cos (x) + 3 tan (x) = 3 sec (x)

14 tan^{2} (x) = - 14 tan (x)

What is the general solution for sin^2(x)=2cos(x)+2,0<= x<= 2pi ?
The general solution for sin^2(x)=2cos(x)+2,0<= x<= 2pi is x=pi