First slide

Trigonometric equations

Question

The least difference between the roots, in the first quadrant $(0 \leq x \leq \frac{π}{2})$ of the equation $4 \cos x (2 - 3 \sin^{2} x) + (\cos 2 x + 1) = 0$ , is

Easy

A

$\frac{π}{6}$
B

$\frac{π}{4}$
C

$\frac{π}{3}$
D

$\frac{π}{2}$

Solution

$W e h a v e 4 \cos x (3 \cos^{2} x - 1) + 2 \cos^{2} x = 0$
$\Rightarrow 2 \cos x (3 \cos x + 2) (2 \cos x - 1) = 0$
$\Rightarrow \cos x = 0 o r \cos x = \frac{1}{2} (∵ \cos x \neq - \frac{2}{3} a s 0 \leq x \leq \frac{π}{2}) \Rightarrow x = \frac{π}{2} o r \frac{π}{3} ∴ t h e d i f f e r e n c e o f v a l u e s i s \frac{π}{2} - \frac{π}{3} = \frac{π}{6}$

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