Trigonometric EquationHard
Question
Number of solutions of $\sqrt{3}cos2\theta + 8cos\theta + 3\sqrt{3} = 0,\theta \in \lbrack - 3\pi,2\pi\rbrack$ is:
Options
A.0
B.5
C.3
D.4
Solution
$\sqrt{3}\left( 2\cos^{2}\theta - 1 \right) + 8cos\theta + 3\sqrt{3} = 0$
$${2\sqrt{3}\cos^{2}\theta + 8cos\theta + 2\sqrt{3} = 0 }{(\sqrt{3}cos + 1)(cos\theta + \sqrt{3}) = 0 }{cos\theta = - \frac{1}{\sqrt{3}} }$$as $- \sqrt{3}$ (reject)
$\therefore\theta =$ will have 5 value in $\lbrack - 3\pi,2\pi\rbrack$
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