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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

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

cos (a) = \frac{1}{2}

sin (2 θ) - cos (2 θ) = 0

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

4 csc (θ) - 5 = 0

tan (4 x) + 3 = 0

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