1. Which of the following is a solution to the equation 2sin^2x+sinx=1
I. x=30°
II. x=150°
III. x=270°
a)III only
b)I and III only
c) II and III only
d)I, II, III
I tried substitution but I dont think it works
Dear ceroseven,
$\displaystyle 2sin^{2}x+sinx=1$
$\displaystyle When, x=30^{o}\Rightarrow{L.H.S=2sin^{2}30^{o}+Sin30^{o} =\left(2\times{\frac{1}{4}}\right)+\frac{1}{2}=1}$
$\displaystyle When, x=150^{o}\Rightarrow{L.H.S=2sin^{2}150^{o}+Sin150^ {o}=2sin^{2}30^{o}+sin30^{o}}$
$\displaystyle Therefore, L.H.S=2sin^{2}150^{o}+Sin150^{o}=\left(2\times{\fr ac{1}{4}}\right)+\frac{1}{2}=1$
$\displaystyle When, x=270^{o}\Rightarrow{L.H.S=2sin^{2}270^{o}+Sin270^ {o}=\left(2\times{-1}\right)+\left(-1\right)=-3}$
Hope this helps.
$\displaystyle 2\sin^2{x} + \sin{x} = 1$
$\displaystyle 2\sin^2{x} + \sin{x} - 1 = 0$
Let $\displaystyle X = \sin{x}$ so that the equation becomes
$\displaystyle 2X^2 + X - 1 = 0$
$\displaystyle 2X^2 + 2X - X - 1 = 0$
$\displaystyle 2X(X + 1) - 1(X + 1) = 0$
$\displaystyle (X + 1)(2X - 1) = 0$
$\displaystyle X + 1 = 0$ or $\displaystyle 2X - 1 = 0$
$\displaystyle X = -1$ or $\displaystyle X = \frac{1}{2}$.
So $\displaystyle \sin{x} = -1$ or $\displaystyle \sin{x} = \frac{1}{2}$
$\displaystyle x = 270^\circ + 360^\circ n$ or $\displaystyle x = \left\{ 30^\circ, 150^\circ \right\} + 360^\circ n$, where $\displaystyle n$ represents the number of times you have gone around the unit circle.
Can you see which solutions fit now?