What is the general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi ?

The general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi is x=0,x=pi,x=2pi

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2sin^2(x)3=7sin(x),0<= x<= 2pi

2 sin^{2} (x) 3 = 7 sin (x), 0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

Solution steps

2 sin^{2} (x) \cdot 3 = 7 sin (x), 0 \leq x \leq 2 π

Solve by substitution

sin (x) = 0, sin (x) = \frac{7}{6}

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = \frac{7}{6}, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = 0, x = π, x = 2 π

6 sin (x) + 12 = 15

cot^{2} (x) cos (x) = cos (x)

\frac{6}{π} - 4 sin^{2} (x) = 0

1 + 2 cos (x) + cos^{2} (x) = 3 sin^{2} (x)

L, sin (k L) = 0

What is the general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi ?
The general solution for 2sin^2(x)3=7sin(x),0<= x<= 2pi is x=0,x=pi,x=2pi