Question

4 cos x - 2 sec x =0

Answer 1

$\bf sec(\theta)=\cfrac{1}{cos(\theta)}\\\\ -------------------------------\\\\ 4cos(x)-2sec(x)=0\implies 4cos(x)-2\cfrac{1}{cos(x)}=0 \\\\\\ 4cos(x)-\cfrac{2}{cos(x)}=0\implies \cfrac{4cos^2(x)-2}{cos(x)}=0 \\\\\\ 4cos^2(x)-2=0\implies 4cos^2(x)=2\implies cos^2(x)=\cfrac{2}{4}$

$\bf cos^2(x)=\cfrac{1}{2}\implies cos(x)=\pm\sqrt{\cfrac{1}{2}}\implies cos(x)=\pm\cfrac{\sqrt{1}}{\sqrt{2}} \\\\\\ cos(x)=\pm\cfrac{1}{\sqrt{2}}\impliedby \textit{and rationalizing the denominator} \\\\\\ cos(x)=\pm\cfrac{\sqrt{2}}{2}\implies \measuredangle x= \begin{cases} \frac{\pi }{4}\\\\ \frac{3\pi }{4}\\\\ \frac{5\pi }{4}\\\\ \frac{7\pi }{4} \end{cases}$

Answer 2

4 cos x - 2 sec x =0 . Note that sec x = 1/(cos x)
4 cos x - 2(1/cos x) = 0  → 4 cos x -2/cos x = 0
Same denominator:

(4.cos x).cos x - 2 = 0

4.cos² x = 2
cos² x = 1/2  → cos x = √1/2 → cos x = 1/√2 = (√2)/√2.√2 = (√2)/2

cos (√2/2) = 45° or π/4