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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

If the sum of the roots of the equation 4 x+6=7 2 x in the interval...

If the sum of the roots of the equation cos⁡4x+6=7cos⁡2x\cos 4 x+6=7 \cos 2 x in the interval [0,314][0,314] is kπ,k∈Rk \pi, k \in R. Find k .

Correct answer

4950

Step-by-step solution

cos⁡4x+6=7cos⁡2x2cos⁡22x−1+6=7cos⁡2x2cos⁡22x−7cos⁡2x+5=02cos⁡22x−2cos⁡2x−5cos⁡2x+5=02cos⁡2x(cos⁡2x−1)−5(cos⁡2x−1)=0(cos⁡2x−1)(2cos⁡2x−5)=0\begin{aligned} & \cos 4 x+6=7 \cos 2 x \\ & 2 \cos ^2 2 x-1+6=7 \cos 2 x \\ & 2 \cos ^2 2 x-7 \cos 2 x+5=0 \\ & 2 \cos ^2 2 x-2 \cos 2 x-5 \cos 2 x+5=0 \\ & 2 \cos 2 x(\cos 2 x-1)-5(\cos 2 x-1)=0 \\ & (\cos 2 x-1)(2 \cos 2 x-5)=0\end{aligned} cos⁡2x=52( not possible )cos⁡2x=1\begin{aligned} & \cos 2 x=\frac{5}{2} \quad(\text { not possible }) \\ & \cos 2 x=1\end{aligned} 2x=2nπ⇒x=nπ\begin{aligned} & 2 x=2 n \pi \\ & \Rightarrow \quad x=n \pi\end{aligned} sum in [0,314][0,314] is π[1+2+3+……+100]−100π=5050π−100π=4950π⇒k=4950\pi[1+2+3+\ldots \ldots+100]-100 \pi=5050 \pi-100 \pi=4950 \pi \Rightarrow k=4950
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