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

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

1) cos(2θ)+6sinēθ=4 and
2)cosθ=sinθ

1) cos(2θ)+6sinēθ=4
cos(2θ)-4=-6(1-cosēθ)
cos(2θ)-4=-6+cosēθ
-cosēθ+cos2θ=-2
-cosēθ+cos2θ+2=0
-cosēθ+2cosēθ+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-cosēθ)
cosθ+cosēθ=1
and I'm stuck there.

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

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

1) cos(2θ)+6sinēθ=4
cos(2θ)-4=-6(1-cosēθ)
cos(2θ)-4=-6+6cosēθ
-cosēθ+cos2θ=-2
-cosēθ+cos2θ+2=0
-cosēθ+2cosēθ+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-cosēθ)
cosθ+cosēθ=1
and I'm stuck there.

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

$\displaystyle (2cos \theta -1)(2cos \theta +1) = 0$

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