What is the general solution for 2sin^2(x/3)=1,-pi<= ,x<= pi ?

The general solution for 2sin^2(x/3)=1,-pi<= ,x<= pi is No Solution for x\in\mathbb{R}

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2sin^2(x/3)=1,-pi<= ,x<= pi

2 sin^{2} (\frac{x}{3}) = 1, - π \leq, x \leq π

No Solution for x \in R

Solution steps

2 sin^{2} (\frac{x}{3}) = 1, - π, x \leq π

Solve by substitution

sin (\frac{x}{3}) = \frac{1}{2}, sin (\frac{x}{3}) = - \frac{1}{2}

sin (\frac{x}{3}) = \frac{1}{2}, - π :

No Solution

sin (\frac{x}{3}) = - \frac{1}{2}, - π :

No Solution

Combine all the solutions

No Solution for x \in R

cos^{2} (x) + sin^{2} (x) = sin (x) + \frac{1}{2}

sin (A) = 0.53

tan (a) = 3, cos (a)

2 sin (2 x) = cos (2 x)

sin (2 θ) = 3

What is the general solution for 2sin^2(x/3)=1,-pi<= ,x<= pi ?
The general solution for 2sin^2(x/3)=1,-pi<= ,x<= pi is No Solution for x\in\mathbb{R}