Question

Solve equation on the interval [0,2pi)

1. 15cos²x - 7 cos x - 2 = 0

Answer 1

`15\cos^2(x)-7\cos(x)-2=0\hspace{5em}\stackrel{\textit{let's just for a few seconds}}{\cos(x)=Z} \\\\\\ 15Z^2-7Z-2=0\implies (5Z+1)(3Z-2)=0 \\\\[-0.35em] ~\dotfill\\\\ 5Z+1=0\implies 5Z=-1\implies Z=-\cfrac{1}{5}\implies \stackrel{\textit{substituting back}}{\cos(x)=-\cfrac{1}{5}} \\\\\\ x=\cos^(-1)\left( -\cfrac{1}{5} \right)\implies x\approx \begin{cases} 101.54^o\\ 258.46^o \end{cases} \\\\[-0.35em] ~\dotfill`

`3Z-2=0\implies 3Z=2\implies Z=\cfrac{2}{3}\implies \stackrel{\textit{substituting back}}{\cos(x)=\cfrac{2}{3}} \\\\\\ x=\cos^(-1)\left( \cfrac{2}{3} \right)\implies x\approx \begin{cases} 48.19^o\\ 311.81^o \end{cases}`

now, how did we get the 2nd angle in each time?

well, let's recall that cos⁻¹(x) will give us an angle only between 0 and π, so 101.54° is on the II Quadrant, however the cosine is also negative on the III Quadrant, we use the reference angle from the II Quadrant, that angle is 180 - 101.54 or 78.46°, so we add that to 180° which is where the III Quadrant starts and we get 180 + 78.46 = 258.46° or very close to that.

we do the same for the 48.19°, that's an angle on the I Quadrant, using that reference angle for the IV Quadrant, we get 360 - 48.19 or about 311.81°.