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

The general solution for -sin(x)(2+sin(x))-cos^2(x)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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

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

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solution steps

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

Rewrite using trig identities

- 1 - 2 sin (x) = 0

Move

1

to the right side

- 2 sin (x) = 1

Divide both sides by

- 2

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

General solutions for

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

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

cos (b) = 0.3125

sin (\frac{1}{2} x) - 1 = 0

- \frac{1}{16} sin (\frac{x}{4}) = 0

sec (x) = - 2, π < x < \frac{3 π}{2}

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

What is the general solution for -sin(x)(2+sin(x))-cos^2(x)=0 ?
The general solution for -sin(x)(2+sin(x))-cos^2(x)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin