What is the general solution for sin^2(x)-3=0 ?

The general solution for sin^2(x)-3=0 is No Solution for x\in\mathbb{R}

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sin^2(x)-3=0

sin^{2} (x) - 3 = 0

No Solution for x \in R

Solution steps

sin^{2} (x) - 3 = 0

Solve by substitution

sin^{2} (x) - 3 = 0

Let:

sin (x) = u

u^{2} - 3 = 0

u^{2} - 3 = 0 : u = 3, u = - 3

u^{2} - 3 = 0

Move

3

to the right side

u^{2} - 3 = 0

Add

3

to both sides

u^{2} - 3 + 3 = 0 + 3

Simplify

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 3, u = - 3

Substitute back

u = sin (x)

sin (x) = 3, sin (x) = - 3

sin (x) = 3, sin (x) = - 3

sin (x) = 3 :

No Solution

sin (x) = 3

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - 3 :

No Solution

sin (x) = - 3

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

cos (θ) csc (θ) = - 2 cos (θ)

csc^{2} (x) - 4 cot (x) = - 2

sin (2 x - \frac{π}{3}) = \frac{3}{2}

- sin^{3} (t) + sin (2 t) cos (t) = 0

tan (θ) = 3 tan (4 5^{\circ})

What is the general solution for sin^2(x)-3=0 ?
The general solution for sin^2(x)-3=0 is No Solution for x\in\mathbb{R}