Trigonometric Equations Question 5
Question 5 - 2024 (29 Jan Shift 2)
The sum of the solutions $x \in \mathbb{R}$ of the equation $\frac{3 \cos 2 x+\cos ^{3} 2 x}{\cos ^{6} x-\sin ^{6} x}=x^{3}-x^{2}+6$ is
(1) 0
(2) 1
(3) -1
(4) 3
Show Answer
Answer (3)
Solution
$\frac{3 \cos 2 x+\cos ^{3} 2 x}{\cos ^{6} x-\sin ^{6} x}=x^{3}-x^{2}+6$
$\Rightarrow \frac{\cos 2 x\left(3+\cos ^{2} 2 x\right)}{\cos 2 x\left(1-\sin ^{2} x \cos ^{2} x\right)}=x^{3}-x^{2}+6$
$\Rightarrow \frac{4\left(3+\cos ^{2} 2 x\right)}{\left(4-\sin ^{2} 2 x\right)}=x^{3}-x^{2}+6$
$\Rightarrow \frac{4\left(3+\cos ^{2} 2 x\right)}{\left(3+\cos ^{2} 2 x\right)}=x^{3}-x^{2}+6$
$x^{3}-x^{2}+2=0 \Rightarrow(x+1)\left(x^{2}-2 x+2\right)=0$
so, sum of real solutions $=-1$
