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

The general solution for 4sin^2(x)-3=0,0<= x<2pi is x= pi/3 ,x=(2pi)/3 ,x=(4pi)/3 ,x=(5pi)/3

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

4 sin^{2} (x) - 3 = 0, 0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

Solution steps

4 sin^{2} (x) - 3 = 0, 0 \leq x < 2 π

Solve by substitution

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

sin (x) = \frac{3}{2}, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{2 π}{3}

sin (x) = - \frac{3}{2}, 0 \leq x < 2 π : x = \frac{4 π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = \frac{π}{3}, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{5 π}{3}

2 \cdot 2^{s i n (x)} = 1

5 sec^{2} (x) - 10 = 0

9 \cdot csc (x) = 63

0 = - sin (t)

∣ sin (x) ∣ = a

What is the general solution for 4sin^2(x)-3=0,0<= x<2pi ?
The general solution for 4sin^2(x)-3=0,0<= x<2pi is x= pi/3 ,x=(2pi)/3 ,x=(4pi)/3 ,x=(5pi)/3