Question

For x $\in (-\pi ,\pi )$ then the number of values of x for which the given

equation ($\sqrt{3}sinx+cosx{)}^{\sqrt{(\sqrt{3}sin2x-cos2x+2)}}$ = 4 is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B

2

The given eqaution is

$(\sqrt{3}sinx+cosx{)}^{\sqrt{(\sqrt{3}sin2x-cos2x+2)}}$ = 4

or$\left[2sin\right(x+\frac{\pi }{6}){]}^{\sqrt{(3si{n}^{2}x+co{s}^{2}x+2\sqrt{3}sinxcosx)}}$ = 4

or$\left[2sin\right(x+\frac{\pi }{6}){]}^{|\sqrt{3}sinx+cosx|}$ = 4

or$\left[2sin\right(x+\frac{\pi }{6}){]}^{|2sin(x+\frac{\pi }{6})|}$ = 4

Hence, 2sin(x + $\frac{\pi }{6}$) = ± 2 or, sin(x + $\frac{\pi }{6}$) = ± 1

or x + $\frac{\pi }{6}$ = n$\pi$ ± $\frac{\pi }{2}$

or x = n$\pi$ ± $\frac{\pi }{2}$ - $\frac{\pi }{6}$

at n = 0, 1,

$\therefore$ x = $\frac{\pi }{3}$ and x = - $\frac{2\pi }{3}$ are the solutions of the given equation.