# Relatively confused...very confused. Solving trig equations, work shown.

• January 10th 2010, 08:52 AM
Voluntarius Disco
Relatively confused...very confused. Solving trig equations, work shown.
1) cos(2θ)+6sin²θ=4 and
2)cosθ=sinθ

1) cos(2θ)+6sin²θ=4
cos(2θ)-4=-6(1-cos²θ)
cos(2θ)-4=-6+cos²θ
-cos²θ+cos2θ=-2
-cos²θ+cos2θ+2=0
-cos²θ+2cos²θ+1=0
cosθ(-cosθ+2cosθ)+1=0
-cosθ+2cosθ+1=0
cosθ+1=0
cosθ=-1
and, as you can imagine, that's not the answer in the back of the book :P
2) cosθ=sinθ
cosθ=(1-cos²θ)
cosθ+cos²θ=1
and I'm stuck there.

Any help would be appreciated - I'm having considerable trouble with these (Headbang)
• January 10th 2010, 08:59 AM
e^(i*pi)
Quote:

Originally Posted by Voluntarius Disco
1) cos(2θ)+6sin²θ=4 and
2)cosθ=sinθ

1) cos(2θ)+6sin²θ=4
cos(2θ)-4=-6(1-cos²θ)
cos(2θ)-4=-6+6cos²θ
-cos²θ+cos2θ=-2
-cos²θ+cos2θ+2=0
-cos²θ+2cos²θ+1=0
cosθ(-cosθ+2cosθ)+1=0

-cosθ+2cosθ+1=0
cosθ+1=0
cosθ=-1
and, as you can imagine, that's not the answer in the back of the book :P
2) cosθ=sinθ
cosθ=(1-cos²θ)
cosθ+cos²θ=1
and I'm stuck there.

Any help would be appreciated - I'm having considerable trouble with these (Headbang)

2) Use the fact that $\frac{sin \theta}{cos \theta} = tan \theta$

Also note that $sin(x) \neq 1-cos^2(x)$ but rather $sin(x) = \sqrt{1-cos^2(x)}$

1) From where you were last correct

$cos(2\theta)-4 = -6+6cos^2 \theta$

$2cos^2\theta-5 = -6+6cos^2\theta$

$4cos^2 \theta -1 = 0$

Use the difference of two squares

$(2cos \theta -1)(2cos \theta +1) = 0$
• January 10th 2010, 09:45 AM
Soroban
Hello, Voluntarius Disco!

Quote:

$1)\;\cos2\theta + 6\sin^2\theta \:=\:4$

. . $\underbrace{\cos2\theta} + 6\sin^2\theta \:=\:4$

$\overbrace{1-2\sin^2\theta} + 6\sin^2\theta \;=\;4 \quad\Rightarrow\quad 4\sin^2\!\theta \:=\:3 \quad\Rightarrow\quad \sin^2\!\theta \:=\:\frac{3}{4}$

. . $\sin\theta \;=\;\pm\frac{\sqrt{3}}{2} \quad\Rightarrow\quad \theta \;=\;\begin{Bmatrix}\dfrac{\pi}{3} + \pi n \\ \\[-3mm] \dfrac{\pi}{3} + \pi n \end{Bmatrix}$

Quote:

$2)\;\cos\theta \:=\:\sin\theta$

We have: . $\sin\theta \;=\;\cos\theta$

Divide by $\cos\theta\!:\;\;\frac{\sin\theta}{\cos\theta} \;=\;\frac{\cos\theta}{\cos\theta} \quad\Rightarrow\quad \tan\theta \;=\;1$

Therefore: . $\theta \;=\;\frac{\pi}{4} + \pi n$