What is the general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi ?

The general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi is x=(3pi)/2

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2sin^2(x)=4sin(x)+6,0<= x<= 2pi

2 sin^{2} (x) = 4 sin (x) + 6, 0 \leq x \leq 2 π

x = \frac{3 π}{2}

Solution steps

2 sin^{2} (x) = 4 sin (x) + 6, 0 \leq x \leq 2 π

Solve by substitution

sin (x) = 3, sin (x) = - 1

sin (x) = 3, 0 \leq x \leq 2 π :

No Solution

sin (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

Combine all the solutions

x = \frac{3 π}{2}

1.33 \cdot sin (6 0^{\circ}) = 1.52 \cdot sin (x)

tan (x) = - 3 cos (x) tan (x)

tan (x - \frac{π}{4}) = \frac{1}{6}

csc (2 χ) = \frac{1}{1 - cos ( 2 χ )}

cos (2 x) - 2 sin (x) = 1

What is the general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi ?
The general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi is x=(3pi)/2