First slide

Trigonometric equations

Question

A solution of the equation $(1 - \tan θ) (1 + \tan θ) \sec^{2} θ + 2^{\tan^{2} θ} = 0$ which $θ$ lies in the interval $(- π / 2, π / 2)$ is given by

Difficult

A

$θ = 0$
B

$θ = \frac{π}{3}$
C

$θ = - \frac{π}{3}$
D

$θ = \frac{π}{6}$

Solution

$\begin{array}{l} (1 - \tan θ) (1 + \tan θ) \sec^{2} θ + 2^{\tan^{2} θ} = 0 \\ \Rightarrow (1 - \tan^{4} θ) + 2^{\tan^{2} θ} = 0 \\ \Rightarrow 2^{t} = t^{2} - 1 (where t = \tan^{2} θ) \end{array}$
By inspection (or by graph) we find
$\begin{array}{l} y = 2^{t} and y = t^{2} - 1 intersect in t = 3 \\ \Rightarrow \tan^{2} θ = 3 \\ \Rightarrow \tan θ = \pm \sqrt{3} \Rightarrow θ = \pm π / 3 \end{array}$

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