Question

The number of solutions of the equation $3(sinx+cosx)-2(si{n}^{3}x+co{s}^{3}x)=8$, $\theta$$ϵ[0,\frac{\pi }{2}]$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$3(\sin x+\cos x)-2({\sin }^{3}x+{\cos }^{3}x)=8$

where $x\in [0,\frac{\pi }{2}]$

or, $3(\sin x+\cos x)-2\left[\right(\sin x+\cos x\left)\right({\sin }^{2}x-\sin x\cos x+{\cos }^{2}x\left)\right]=8$

or, $3(\sin x+\cos x)-2\left[\right(\sin x+\cos x\left)\right(1-\sin x\cos x\left)\right]=8$

or, $3(\sin x+\cos x)-[3-2+2\sin x\cos x]=8$

or, $3(\sin x+\cos x)-[1+2\sin x\cos x]=8$

or, $3(\sin x+\cos x)-2[{\sin }^{2}x+2\sin x\cos x]+{\cos }^{2}x]=8$

or, (\sin x+\cos x)^{3}=8$

or, $\sin x+\cos x=2$

This is possible if $\sin x=1$ and $\cos x=1$.

But both the values cannot be 1 at the same time.