Aspire Faculty ID #16430 · Topic: NIMCET 2009 · Just now
NIMCET 2009

The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for:

Solution

Simplify: $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$ 
$= 1 - \dfrac{1}{2}\sin^2(2x)$ 
So equation becomes: 
$1 - \dfrac{1}{2}\sin^2(2x) + \sin(2x) + \alpha = 0$ 
Let $t = \sin(2x)$, where $t \in [-1,1]$ 
Expression becomes: $1 + \alpha + t - \dfrac{t^2}{2} = 0$ 
Multiply by 2: 
$t^2 - 2t - 2(1+\alpha) = 0$ 
For real $t$ in $[-1,1]$, discriminant must allow roots inside interval. Solving gives the range: 
${-\dfrac{3}{2} \le \alpha \le \dfrac{1}{2}}$

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