# Thread: 2 trig problems, 0 ideas...

1. ## 2 trig problems, 0 ideas...

Find all values of x in [0,2pi]

sin2xsinx=cosx

and

All values of x:

2cos^(2) x + sinx = 1

I'd appreciate any help! Thanks!

2. Originally Posted by JasonW
Find all values of x in [0,2pi]

sin2xsinx=cosx
For $sin2xsinx = cosx$

Double Angle Identities
$2sinxcosx \times sinx - cosx = 0$

Factor
$cosx(2sin^2x - 1) = 0$

First:
$cosx = 0$(since value is 0, refer to graph)
$x = \frac {\pi}{2} and \frac {3\pi}{2}$

Second:

$2sin^2x - 1 = 0$
Simplify
$sin^2x = 1/2$

$sinx = \pm 1/2$

$x = sin^-1(1/2)$

Remember the CAST Rule
$x = \frac {\pi}{6} and \frac {5\pi}{6}$

$x = sin^-1(-1/2)$

$x = \frac {11\pi}{6} and \frac {7\pi}{6}$

3. Originally Posted by JasonW

and

All values of x:

2cos^(2) x + sinx = 1

I'd appreciate any help! Thanks!
Move everything to one side
$2cos^2x + sinx = 1$

Make everything in one term, only sin or only cos
$2(1 - sin^2x) + sinx - 1 = 0$

Move everything to the right side since you want your constant number for the first term to be positive
$2 - 2sin^2x + sinx - 1 = 0$

$0 = 2sin^2x - sinx - 1$

This is a complex trinomial, so you need to use decomposition
$0 = 2sin^2x - 2sinx + sinx - 1$

Simplify the trinomial
$0 = 2sinx (sinx - 1) + 1 (sinx - 1)$

Set each to zero by themselves
$0 = (sinx - 1) (2sinx + 1)$

Since sinx = 1, you can find the x values by graph
$sinx - 1 =0$

$sinx = 1$

$x = \frac {\pi}{2}$

Determine which quadrant the angles are going to be in (CAST Rule), since the value of sin is negative, it'll be in Q3 and Q4
$2sinx + 1 =0$

$sinx = \frac {-1}{2}$

$x = sin^-1 \frac {-1}{2}$

Since there is a negative degrees (in negative coterminal), you need to find the related acute angle which is 30

$x = -30$

$x = \frac {11\pi}{6} and \frac {7\pi}{6}$