What is the general solution for sin^2(θ)-16=0 ?

The general solution for sin^2(θ)-16=0 is No Solution for θ\in\mathbb{R}

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sin^2(θ)-16=0

sin^{2} (θ) - 16 = 0

No Solution for θ \in R

Solution steps

sin^{2} (θ) - 16 = 0

Solve by substitution

sin^{2} (θ) - 16 = 0

Let:

sin (θ) = u

u^{2} - 16 = 0

u^{2} - 16 = 0 : u = 4, u = - 4

u^{2} - 16 = 0

Move

16

to the right side

u^{2} - 16 = 0

Add

16

to both sides

u^{2} - 16 + 16 = 0 + 16

Simplify

u^{2} = 16

u^{2} = 16

For

x^{2} = f (a)

the solutions are

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

u = 16, u = - 16

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

- 16 = - 4

- 16

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= - 4

u = 4, u = - 4

Substitute back

u = sin (θ)

sin (θ) = 4, sin (θ) = - 4

sin (θ) = 4, sin (θ) = - 4

sin (θ) = 4 :

No Solution

sin (θ) = 4

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = - 4 :

No Solution

sin (θ) = - 4

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for θ \in R

sin (x) + tan (x) = 0

sin^{2} (x) cos^{2} (x) = 0

3 sin (x) cos (x) + 5 cos (x) = 0

5 sin (x) tan (x) = - 7 tan (x)

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

What is the general solution for sin^2(θ)-16=0 ?
The general solution for sin^2(θ)-16=0 is No Solution for θ\in\mathbb{R}