What is the general solution for 2sin^2(x)-sin(x)-1=0,0<= x<= 360 ?

The general solution for 2sin^2(x)-sin(x)-1=0,0<= x<= 360 is x=90,x=210,x=330

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2sin^2(x)-sin(x)-1=0,0<= x<= 360

2 sin^{2} (x) - sin (x) - 1 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

x = 9 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}

Solution steps

2 sin^{2} (x) - sin (x) - 1 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

Solve by substitution

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

sin (x) = 1, 0 \leq x \leq 36 0^{\circ} : x = 9 0^{\circ}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 36 0^{\circ} : x = 21 0^{\circ}, x = 33 0^{\circ}

Combine all the solutions

x = 9 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}

x, z = sin (x^{2} y)

2 tan (x) - 3 = 3 tan (x) - 4

1 + cot (θ) = csc (θ)

tan (x) = cot (2 5^{\circ})

3 csc^{2} (5 x) = 4

What is the general solution for 2sin^2(x)-sin(x)-1=0,0<= x<= 360 ?
The general solution for 2sin^2(x)-sin(x)-1=0,0<= x<= 360 is x=90,x=210,x=330