Question

Find the number of solutions of the equations
$(\sin x-1{)}^3+(\cos x-1{)}^3+(\sin x{)}^3=(2\sin x+\cos x-2{)}^3$ in $[0,2\pi ]$

Answer 1

Given,

$(\sin x-1{)}^3+(\cos x-1{)}^3+(\sin x{)}^3=(2\sin x+\cos x-2{)}^3$----------------1

We know that, $(a+b+c{)}^3=a^3+b^3+c^3+(a+b+c\left)\right(ab+bc+ca)$

Here $a=\sin x-1,b=\cos x-1,c=\sin x$

$(a+b+c)=(2\sin x+\cos x-2)$

$(ab+bc+ca)=(2{\sin }^{2}x+2\sin x\cos x-3\sin x-\cos x+1)$

(1)$(\sin x-1{)}^3+(\cos x-1{)}^3+(\sin x{)}^3=(\sin x-1{)}^3+(\cos x-1{)}^3+(\sin x{)}^3+(2\sin x+\cos x-2)(2{\sin }^{2}x+2\sin x\cos x-3\sin x-\cos x+1)$

$\Rightarrow (2\sin x+\cos x-2)(2{\sin }^{2}x+2\sin x\cos x-2\sin x-\sin x-\cos x+1)=0$

$\Rightarrow (2\sin x+\cos x-2)(2\sin x-1)(\sin x+\cos x-1)=0$

(i) $\sin x=\frac{1}{2}$

$\therefore x=\frac{\pi }{6},\frac{5\pi }{6}$ ( as $x\in [0,2\pi ]$)

or

(ii) $\sin x+\cos x=1$

$\therefore x=0,\frac{\pi }{2},2\pi$ ( as $x\in [0,2\pi ])$

or

(iii) $2\sin x+\cos x-2=0$

$2(\sin x-1)=-\cos x$

$4{\sin }^{2}x-8\sin x+4={\cos }^{2}x$

$5{\sin }^{2}x-8\sin x+3=0$

$\therefore \sin x=1or\frac{3}{5}$

$\cos x=2(1-\sin x)$

$=2(1-\frac{3}{5})=\frac{4}{5}$

$\sin x>0$ and $\cos x>0$

$\therefore$ x has one solution in 1st quadrant.

From (i),(ii),(iii), number of solutions of $x=6$