Answer:
[tex]x=\dfrac{\pi}{6},\dfrac{5\pi}{6}[/tex]
Step-by-step explanation:
Recalling the secant/tangent Pythagorean identity, [tex]\rm1+tan^2x=sec^2x[/tex] we can see its rearranged version on the right side of our equation.
[tex]\rm1+tan^2x=sec^2x \rightarrow1=sec^2x-tan^2x[/tex]
[tex]\rm4cos^2x-2=1[/tex]
[tex]\dotfill[/tex]
All there's left is to isolate the cosine term and find values of x that satisfy the cosine equation.
[tex]\rm 4cos^2x=3[/tex]
[tex]\rm cos^2x=\dfrac{3}{4}[/tex]
[tex]\rm cosx=\pm\sqrt{\dfrac{3}{4} } =\pm\dfrac{\sqrt{3} }{2}[/tex]
Being mindful of the restriction, [tex]x=\dfrac{\pi}{6} ,\dfrac{5\pi}{6} ,\dfrac{7\pi}{6} ,\dfrac{11\pi}{6}[/tex]. Since the problem wants the answers from least to greatest and has two spots we consider the first two answers: [tex]x=\dfrac{\pi}{6},\dfrac{5\pi}{6}[/tex].