What is the general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi ?

The general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi is x=pi

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cos(x)=-sin^2(x)-1,0<= x<= 2pi

cos (x) = - sin^{2} (x) - 1, 0 \leq x \leq 2 π

x = π

Solution steps

cos (x) = - sin^{2} (x) - 1, 0 \leq x \leq 2 π

Subtract

- sin^{2} (x) - 1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

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

cos (x) = 2, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = π

θ, cos (θ) = \frac{- 1 2}{2}

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

sin (2 x) = cos (x - 1 5^{\circ})

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

arcsin (\frac{3}{5}) + arccos (x) = π

What is the general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi ?
The general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi is x=pi