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

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

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

- 2 sin (x) - 2 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

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

Rewrite using trig identities

- (1 - 2 sin^{2} (x)) \cdot 2 - 2 sin (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

\frac{cos ( x )}{1 + sin ( x )} = 1 - \frac{sin ( x )}{cos ( x )}

y, 2 cos (5 x + y) = 0

2 cos^{2} (x) + 15 sin (x) - 15 = 0

cos (2 x) = sin (x), - 2 π \leq x \leq 2 π

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

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