JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q88

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MathematicsSingle correct+4 / -1Question 88 of 90

If $5(\tan^2x-\cos^2x)=2\cos2x+9$ , then the value of $\cos4x$ is :

\dfrac13

\dfrac29

-\dfrac79

-\dfrac35

Answer and worked solution

Official answer: C

With $c=\cos^2x$ : $5\left(\dfrac{1-c}c-c\right)=4c+7$ , i.e. $9c^2+12c-5=0$ , so $c=\frac13$ . Then $\cos2x=-\frac13$ and
$\cos4x=2\cdot\frac19-1=-\frac79.$

Topics:Trigonometric equations

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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 88