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

The general solution for solvefor x,0=-sin(2x)-2cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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

solve for

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

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

Switch sides

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

Rewrite using trig identities

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

Factor

- 2 cos (x) - 2 cos (x) sin (x) : - 2 cos (x) (sin (x) + 1)

- 2 cos (x) (sin (x) + 1) = 0

Solving each part separately

cos (x) = 0 or sin (x) + 1 = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) + 1 = 0 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sec (θ) = \frac{5}{2}

4 cos (x - 1) = 0

cos (θ) = \frac{5}{7}, cos (\frac{θ}{2}), 0^{\circ} < θ < 9 0^{\circ}

sin (x) + 4 = cos (x) + 3

\frac{sin ( 17 0 ^{\circ} )}{8} = \frac{sin ( B )}{5}

What is the general solution for solvefor x,0=-sin(2x)-2cos(x) ?
The general solution for solvefor x,0=-sin(2x)-2cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin