1) cos(2θ)+6sin²θ=4 and

2)cosθ=sinθ

1) cos(2θ)+6sin²θ=4

cos(2θ)-4=-6(1-cos²θ)

cos(2θ)-4=-6+

**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

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

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