To follow up:
From the Pythagorean theorem, we find the hypotenuse \(h\) is:
[MATH]h=\sqrt{3^2+4^2}=5[/MATH]
Using a double-angle identity for cosine, we may write:
[MATH]\cos(2\theta)=1-2\sin^2(\theta)=1-2\left(\frac{-3}{5}\right)^2=\frac{7}{25}[/MATH]
For the 2nd problem, we can write as a quadratic in \(\cos(x)\) in standard form:
[MATH]5\cos^2(x)+4\cos(x)-1=0[/MATH]
Factor:
[MATH](5\cos(x)-1)(\cos(x)+1)=0[/MATH]
From the first factor, we find:
[MATH]x\in\left\{\arccos\left(\frac{1}{5}\right),2\pi-\arccos\left(\frac{1}{5}\right)\right\}[/MATH]
And from the second factor we find:
[MATH]x\in\{\pi\}[/MATH]