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JEE Main 2024
Trigonometric Equations
Trigonometric Equations
Hard

Question

Let S={sin22θ:(sin4θ+cos4θ)x2+(sin2θ)x+(sin6θ+cos6θ)=0S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\left(\sin ^6 \theta+\cos ^6 \theta\right)=0\right. has real roots }\}. If α\alpha and β\beta be the smallest and largest elements of the set SS, respectively, then 3((α2)2+(β1)2)3\left((\alpha-2)^2+(\beta-1)^2\right) equals __________.

Answer: 0

Solution

For real roots D0sin22θ4(sin4θ+cos4θ)(sin6θ+cos6θ)\begin{aligned} & D \geq 0 \\ & \sin ^2 2 \theta \geq 4\left(\sin ^4 \theta+\cos ^4 \theta\right)\left(\sin ^6 \theta+\cos ^6 \theta\right) \end{aligned} Put sin22θ=t\sin ^2 2 \theta=t t4(1t2)(13t4)2t(2t)(43t)3t212t+80t24t+830(t2)2+8340(t2)24323t223223t2+23t[0,1]223t1α=223,β=13[(α2)2+(β1)2]=4\begin{aligned} & \Rightarrow t \geq 4\left(1-\frac{t}{2}\right)\left(1-\frac{3 t}{4}\right) \\ & 2 t \geq(2-t)(4-3 t) \\ & 3 t^2-12 t+8 \leq 0 \\ & t^2-4 t+\frac{8}{3} \leq 0 \\ & (t-2)^2+\frac{8}{3}-4 \leq 0 \\ & (t-2)^2 \leq \frac{4}{3} \\ & -\frac{2}{\sqrt{3}} \leq t-2 \leq \frac{2}{\sqrt{3}} \\ & 2-\frac{2}{\sqrt{3}} \leq t \leq 2+\frac{2}{\sqrt{3}} \\ & \because t \in[0,1] \\ & \Rightarrow 2-\frac{2}{\sqrt{3}} \leq t \leq 1 \\ & \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \\ & \Rightarrow 3\left[(\alpha-2)^2+(\beta-1)^2\right]=4 \end{aligned}

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