# trig equations

• Feb 3rd 2010, 11:02 PM
ceroseven
trig equations
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
• Feb 4th 2010, 12:25 AM
Sudharaka
Dear ceroseven,

$2sin^{2}x+sinx=1$

$When, x=30^{o}\Rightarrow{L.H.S=2sin^{2}30^{o}+Sin30^{o} =\left(2\times{\frac{1}{4}}\right)+\frac{1}{2}=1}$

$When, x=150^{o}\Rightarrow{L.H.S=2sin^{2}150^{o}+Sin150^ {o}=2sin^{2}30^{o}+sin30^{o}}$

$Therefore, L.H.S=2sin^{2}150^{o}+Sin150^{o}=\left(2\times{\fr ac{1}{4}}\right)+\frac{1}{2}=1$

$When, x=270^{o}\Rightarrow{L.H.S=2sin^{2}270^{o}+Sin270^ {o}=\left(2\times{-1}\right)+\left(-1\right)=-3}$

Hope this helps.
• Feb 4th 2010, 01:46 AM
Prove It
Quote:

Originally Posted by ceroseven
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

$2\sin^2{x} + \sin{x} = 1$

$2\sin^2{x} + \sin{x} - 1 = 0$

Let $X = \sin{x}$ so that the equation becomes

$2X^2 + X - 1 = 0$

$2X^2 + 2X - X - 1 = 0$

$2X(X + 1) - 1(X + 1) = 0$

$(X + 1)(2X - 1) = 0$

$X + 1 = 0$ or $2X - 1 = 0$

$X = -1$ or $X = \frac{1}{2}$.

So $\sin{x} = -1$ or $\sin{x} = \frac{1}{2}$

$x = 270^\circ + 360^\circ n$ or $x = \left\{ 30^\circ, 150^\circ \right\} + 360^\circ n$, where $n$ represents the number of times you have gone around the unit circle.

Can you see which solutions fit now?