1. ## Solving Trig Equations

Hi guys,
I have 2 equations on my homework that I am not allowed to use a calculator on, and need to find the exact answer. Sorry im posting so much, I got a gigantic packet to work on.

$\displaystyle 2sin^{2}(x)+5cos(x)=4$where$\displaystyle 0 \le x\le360$

AND

$\displaystyle sin(2x)=2cos^{2}(x)$where$\displaystyle 0\le x\le 2PI$

2. Originally Posted by spycrab
Hi guys,
I have 2 equations on my homework that I am not allowed to use a calculator on, and need to find the exact answer. Sorry im posting so much, I got a gigantic packet to work on.

$\displaystyle 2sin^{2}(x)+5cos(x)=4$where$\displaystyle 0 \le x\le360$

AND

$\displaystyle sin(2x)=2cos^{2}(x)$where$\displaystyle 0\le x\le 2PI$
for the first equation, change $\displaystyle \sin^2{x}$ to $\displaystyle 1-\cos^2{x}$ , move everything to one side, and factor the resulting quadratic.

for the second, use the double angle identity for sine, move everything to one side, and factor.

3. I got x = 60 for the first problem, but for the second, when I
use the double angle identity i get
$\displaystyle 2sin(x)cos(x)=2cos^{2}(x)$
i move it
$\displaystyle 0=2cos^{2}(x)-2sin(x)cos(x)$
I factor cos(x) out
$\displaystyle 0=cos(x)(2cos(x)-2sin(x)$
i factor the 2 out
$\displaystyle 2cos(x)(cos(x)-sin(x))$
but then I dont know what to do

4. Originally Posted by spycrab
I got x = 60 for the first problem, but for the second, when I
use the double angle identity i get
$\displaystyle 2sin(x)cos(x)=2cos^{2}(x)$
i move it
$\displaystyle 0=2cos^{2}(x)-2sin(x)cos(x)$
I factor cos(x) out
$\displaystyle 0=cos(x)(2cos(x)-2sin(x)$
i factor the 2 out
$\displaystyle 2cos(x)(cos(x)-sin(x))$
but then I dont know what to do
set each factor equal to 0 ...

$\displaystyle 2\cos{x} = 0$

$\displaystyle \cos{x} - \sin{x} = 0$

solve each.

5. im having problems with cosx-sinx = 0. for the 2cosx = 0 part, i get pi/2,

6. Originally Posted by spycrab
im having problems with cosx-sinx = 0. for the 2cosx = 0 part, i get pi/2,
$\displaystyle \cos{x} = 0$ has two solutions

$\displaystyle \cos{x} = \sin{x}$ has two solutions

look at the unit circle ...

7. Originally Posted by spycrab
im having problems with cosx-sinx = 0. for the 2cosx = 0 part, i get pi/2,
$\displaystyle Cos(x)=Sin(x)$

For what angles are the horizontal (Cos(x)) and vertical (Sin(x)) co-ordinates of a point on the unit-circle identical ?

Also, there is a second solution for 2Cosx=0.

8. Originally Posted by spycrab
I got x = 60 for the first problem
you're missing a solution

9. Thanks!

I got $\displaystyle x=\frac{\pi}{2},x=\frac{\pi}{4},x=\frac{5\pi}{4},x =\frac{3\pi}{2}$