Question

The general solution of trigonometric equation $1+si{n}^{3}x+co{s}^{3}x=\frac{3}{2}sin2x$ is

• A. $2n\pi \pm \frac{3\pi }{4}+\frac{\pi }{4}$
• B. $2n\pi \pm \frac{\pi }{4}+\frac{3\pi }{4}$
• C. $2n\pi \pm \frac{\pi }{4}$
• D. $2n\pi \pm \frac{3\pi }{4}$

Answer 1

The correct option is A $2n\pi \pm \frac{3\pi }{4}+\frac{\pi }{4}$

If a+b + c= 0 then $a^3+b^3+c^3=3abc$
$1+si{n}^{3}x+co{s}^{3}x=\frac{3}{2}sin2x\Rightarrow 1+sinx+cosx=0$
$\Rightarrow sinx+cosx=-1$
$\Rightarrow cos(x-\frac{\pi }{4})=\frac{-1}{2}$
$\therefore x-\frac{\pi }{4}=2n\pi \pm \frac{3\pi }{4}$
$\Rightarrow x=2n\pi \pm \frac{3\pi }{4}+\frac{\pi }{4}$