What is the general solution for -2/3 cos(x)+2/3 cos(2x)=0 ?

The general solution for -2/3 cos(x)+2/3 cos(2x)=0 is x=2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin

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-2/3 cos(x)+2/3 cos(2x)=0

- \frac{2}{3} cos (x) + \frac{2}{3} cos (2 x) = 0

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solution steps

- \frac{2}{3} cos (x) + \frac{2}{3} cos (2 x) = 0

Simplify

- \frac{2}{3} cos (x) + \frac{2}{3} cos (2 x) : \frac{- 2 cos ( x ) + 2 cos ( 2 x )}{3}

\frac{- 2 cos ( x ) + 2 cos ( 2 x )}{3} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

(- 1 + 2 cos^{2} (x)) \cdot 2 - 2 cos (x) = 0

Solve by substitution

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

cos (x) = 1 : x = 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sin (θ) - cos (θ) = 0

cot (3 x) = - 3, 0 \leq x \leq 2 π

3 cos^{2} (x) + sin^{2} (x) = 3

4 sin^{2} (x) + 6 sin (x) + 2 = 0

sin^{2} (θ) + 2 - cos^{2} (θ) = 3 sin (θ)

What is the general solution for -2/3 cos(x)+2/3 cos(2x)=0 ?
The general solution for -2/3 cos(x)+2/3 cos(2x)=0 is x=2pin,x=(2pi)/3+2pin,x=(4pi)/3+2pin