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

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

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

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

Rewrite using trig identities

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

sin (x) - cos (x) = 1, 0 \leq x \leq 2 π

0 = sin (0 + c)

2 + 2 cos (θ) = \frac{2}{1 + cos ( θ )}

14 sin (2 x) + 7 = 0

tan (3 x) + sec (3 x) = 2

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