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

The general solution for 2sin^2(x)-7sin(x)=-3,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6

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

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

x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

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

Solve by substitution

sin (x) = 3, sin (x) = \frac{1}{2}

sin (x) = 3, 0 \leq x \leq 2 π :

No Solution

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}

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

(218) sin^{2} (x) + (126) sin (x) - 88 = 0

3 sin (3 x) = 0

cos (θ) csc (θ) = sin (θ) sec (θ)

64 cos^{2} (θ) = 49, (0^{\circ} \leq, θ \leq 36 0^{\circ})

What is the general solution for 2sin^2(x)-7sin(x)=-3,0<= x<= 2pi ?
The general solution for 2sin^2(x)-7sin(x)=-3,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6