First slide

Trigonometric equations

Question

$Number of solution(s) of the equation sin 2 θ + \cos 2 θ = - \frac{1}{2}, θ \in [0, \frac{π}{2}], is$

Easy

A

0
B

1
C

2
D

3

Solution

$\begin{array}{l} Given that \sin 2 θ + \cos 2 θ = - \frac{1}{2} \\ \Rightarrow \frac{2 \tan θ}{1 + \tan^{2} θ} + \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} = - \frac{1}{2} \\ \Rightarrow \frac{2 \tan θ + 1 - \tan^{2} θ}{1 + \tan^{2} θ} = - \frac{1}{2} \\ \Rightarrow \tan^{2} θ - 4 \tan θ - 3 = 0 \end{array}$
$\begin{array}{l} It is a quadratic equation in terms of \tan θ \\ \Rightarrow \tan θ = \frac{4 \pm \sqrt{16 + 12}}{2} (∵ \tan θ = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}) \end{array}$
$\begin{array}{l} \Rightarrow \tan θ = 2 \pm \sqrt{7} \\ Number of solution of the given equation is 1 \end{array}$ $(\sin c e \tan θ i s p o s i t i v e)$

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