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

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

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

2 sin^{2} (x) - 7 sin (x) + 3 = 0, 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, 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}

csc (θ) = - \frac{( 2 3 )}{3}

tan (x - 2 0^{\circ}) = tan (2 x + 10)

sin (5 x) cos (x) - cos (5 x) sin (x) = 0

2 sin (3 x + 1) + 1 = 2

2 sin (3 x + 1) + 1 = 1

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