What is the general solution for 4sin^2(x)-3=4,sin(2x)-3=0 ?

The general solution for 4sin^2(x)-3=4,sin(2x)-3=0 is No Solution for x\in\mathbb{R}

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4sin^2(x)-3=4,sin(2x)-3=0

4 sin^{2} (x) - 3 = 4, sin (2 x) - 3 = 0

No Solution for x \in R

Solution steps

4 sin^{2} (x) - 3 = 4, sin (2 x) - 3 = 0

Solve by substitution

sin (x) = \frac{7}{2}, sin (x) = - \frac{7}{2}

sin (x) = \frac{7}{2}, sin (2 x) - 3 = 0 :

No Solution

sin (x) = - \frac{7}{2}, sin (2 x) - 3 = 0 :

No Solution

Combine all the solutions

No Solution for x \in R

tan^{2} (θ) + tan (θ) = 6, 0 \leq θ < 18 0^{\circ}

sin (2 x) = 5 sin (x)

2 cos^{2} (θ) - 4 cos (θ) - 1 = - 5 cos (θ)

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

tan (x) = - 3 + 2

What is the general solution for 4sin^2(x)-3=4,sin(2x)-3=0 ?
The general solution for 4sin^2(x)-3=4,sin(2x)-3=0 is No Solution for x\in\mathbb{R}