Answer to Question #164596 in Trigonometry for Samir khan

Question

Solve for x in degrees, where 0 degrees <= x < 360 degrees . 2sin^2 x - sin x = 1

Answer 1

Solution.

$$2\sin^2{x}-\sin{x}=1, \text{ }0^{\degree}\leq x <360^{\degree}.$$

Make a replacement: $t=\sin{x},$ then $2t^2-t-1=0.$

Solve this equation:

$$D=b^2-4ac=1-4\cdot 2\cdot (-1)=9,$$ $$t_1=\frac{-b-\sqrt{D}}{2a}=\frac{1-3}{2\cdot 2}=-\frac{1}{2}; t_2=\frac{-b+\sqrt{D}}{2a}=\frac{1+3}{2\cdot 2}=1.$$

Back to the replacement:

1)

$\sin{x}=-\frac{1}{2}, \newline x=(-1)^n\arcsin{(-\frac{1}{2})}+\pi n, n \in Z, \newline x=(-1)^{n+1}\arcsin{\frac{1}{2}}+\pi n, n \in Z, \newline x=(-1)^{n+1}\frac{\pi}{6}+\pi n, n \in Z, \newline x=(-1)^{n+1}30^{\degree}+180^{\degree} n, n \in Z.$

At

$n=1: x=30^{\degree}+180^{\degree}=210^{\degree}, \newline n=2: x=-30^{\degree}+180^{\degree}\cdot 2=330^{\degree}. \newline$

2)

$\sin{x}=1, \newline x=\frac{\pi}{2}+2\pi k, k \in Z, \newline x=90^{\degree}+360^{\degree}k, k \in Z. \newline$

At

$k=0: x=90^{\degree}.$

Answer. $x=210^{\degree}, x=330^{\degree}, x=90^{\degree}.$