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

Aug 2009
13
0
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)
 
Feb 2009
3,053
1,333
West Midlands, England
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 \(\displaystyle \frac{sin \theta}{cos \theta} = tan \theta\)

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

1) From where you were last correct

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

\(\displaystyle 2cos^2\theta-5 = -6+6cos^2\theta\)

\(\displaystyle 4cos^2 \theta -1 = 0\)

Use the difference of two squares

\(\displaystyle (2cos \theta -1)(2cos \theta +1) = 0\)
 
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Reactions: Voluntarius Disco
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, Voluntarius Disco!

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

. .\(\displaystyle \underbrace{\cos2\theta} + 6\sin^2\theta \:=\:4\)

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

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




\(\displaystyle 2)\;\cos\theta \:=\:\sin\theta\)

We have: .\(\displaystyle \sin\theta \;=\;\cos\theta\)

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

Therefore: .\(\displaystyle \theta \;=\;\frac{\pi}{4} + \pi n\)