Question 7.6.5: Solving a Trigonometric Equation (Squaring) Solve tan x + √3...
Solving a Trigonometric Equation (Squaring)
Solve \tan x + \sqrt{3} = \sec x over the interval [0, 2π).
We must rewrite the equation in terms of a single trigonometric function. Because the tangent and secant functions are related by the identity 1 + \tan² x = \sec² x, square each side and express \sec² x in terms of tan2 x.
(\tan x + \sqrt{3})² = (\sec x)² Square each side.
\fbox{Don’t forget the middle term.}↓
\tan² x + 2\sqrt{3} \tan x + 3 = \sec² x (x + y)² = x² + 2xy + y²
\tan² x + 2\sqrt{3} \tan x + 3 = 1 + \tan² x Pythagorean identity
2\sqrt{3} \tan x = -2 Subtract 3 + \tan² x.
\tan x = – \frac{1}{\sqrt{3}} Divide by 2\sqrt{3}.
\tan x = – \frac{\sqrt{3}}{3} Rationalize the denominator.
Solutions of \tan x = – \frac{\sqrt{3}}{3} over [0, 2π) are \frac{5π}{6} and \frac{11π}{6} . These possible, or proposed, solutions must be checked to determine whether they are also solutions of the original equation.
CHECK \tan x + \sqrt{3} = \sec x Original equation
\left.\begin{matrix} \tan \left(\frac{5\pi }{6} \right)+\sqrt{3}≟\sec\left(\frac{5\pi }{6} \right) \\ \\ -\frac{\sqrt{3} }{3}+\frac{3\sqrt{3} }{3}≟-\frac{2\sqrt{3} }{3} \\ \\ \frac{2\sqrt{3} }{2}=-\frac{2\sqrt{3} }{3}& \text{False} \end{matrix} \right| \begin{matrix} \tan \left(\frac{11\pi }{6} \right)+\sqrt{3}≟\sec \left(\frac{11\pi }{6} \right) \\ \\ -\frac{\sqrt{3} }{3} +\frac{3\sqrt{3} }{3} ≟\frac{2\sqrt{3} }{3} \\ \\ \frac{2\sqrt{3} }{3}=\frac{2\sqrt{3} }{3} ✓ & \text{ True} \end{matrix}As the check shows, only \frac{11\pi }{6} is a solution, so the solution set is { \frac{11π}{6} }.
