Start free

Mathematics · Trigonometry Ratios and Identities

JEE Main 2024 — 29 January, Shift 2 — Question 7

The sum of the solutions x∈Rx \in \mathbb{R} of the equation 3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\frac{3 \cos 2 x+\cos ^{3} 2 x}{\cos ^{6} x-\sin ^{6} x}=x^{3}-x^{2}+6 is

  1. Option A:

    0

  2. Option B:

    1

  3. Option C:

    -1

    Correct
  4. Option D:

    3

Answer: C

Step-by-step solution

3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\frac{3 \cos 2 x+\cos ^{3} 2 x}{\cos ^{6} x-\sin ^{6} x}=x^{3}-x^{2}+6

⇒cos⁡2x(3+cos⁡22x)cos⁡2x(1−sin⁡2xcos⁡2x)=x3−x2+6\Rightarrow \frac{\cos 2 \mathrm{x}\left(3+\cos ^{2} 2 \mathrm{x}\right)}{\cos 2 \mathrm{x}\left(1-\sin ^{2} \mathrm{x} \cos ^{2} \mathrm{x}\right)}=\mathrm{x}^{3}-\mathrm{x}^{2}+6

⇒4(3+cos⁡22x)(4−sin⁡22x)=x3−x2+6\Rightarrow \frac{4\left(3+\cos ^{2} 2 x\right)}{\left(4-\sin ^{2} 2 x\right)}=x^{3}-x^{2}+6

⇒4(3+cos⁡22x)(3+cos⁡22x)=x3−x2+6\Rightarrow \frac{4\left(3+\cos ^{2} 2 x\right)}{\left(3+\cos ^{2} 2 x\right)}=x^{3}-x^{2}+6

x3−x2+2=0⇒(x+1)(x2−2x+2)=0\mathrm{x}^{3}-\mathrm{x}^{2}+2=0 \Rightarrow(\mathrm{x}+1)\left(\mathrm{x}^{2}-2 \mathrm{x}+2\right)=0

So, sum of real solutions =−1=-1

Answer key and solution verified before publishing.

Practise Trigonometry Ratios and Identities

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Periodicity of trigometric functions,Solutions of trigonometric equations
The sum of the solutions x in mathbb R of the equation frac 3 cos 2… | JEE Main 2024 PYQ with Solution · DhiX AI