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

The general solution for -2sin(x)+2cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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

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

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 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{3 π}{2} + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sec^{2} (x) = 0

sec^{2} (x) + tan (x) = 1

4 sin^{2} (θ) - 1 = 0

tan^{2} (x) = tan (x)

cos (x) = - sin (x)

What is the general solution for -2sin(x)+2cos(2x)=0 ?
The general solution for -2sin(x)+2cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin