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Answer:

[tex]x=\dfrac{\pi}{6},\dfrac{5\pi}{6}[/tex]

Step-by-step explanation:

Solving the Problem

Simplifying the Equation

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]

Solving for X

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].

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