IIT-JEE 2005Question 1
Find the range of values of $t$ for which $$2\sin t=\frac{1-2x+5x^{2}}{3x^{2}-2x-1},\quad t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].$$
Trigonometric Ratio & Identities
Answer$t\in\left[-\dfrac{\pi}{2},-\dfrac{\pi}{10}\right]\cup\left[\dfrac{3\pi}{10},\dfrac{\pi}{2}\right]$
Solution
Concept. Put $y=2\sin t$. The equation has a real $x$ only if a quadratic in $x$ has a non-negative discriminant; that bounds $y$. Formulas used. (1) $px^2+qx+r=0$ has real roots iff $q^2-4pr\ge0$(2) $\sin18^\circ=\dfrac{\sqrt5-1}{4}$ and $\sin54^\circ=\cos36^\circ=\dfrac{\sqrt5+1}{4}$Step 1 - Quadratic in $x$. Cross-multiplying, $$(3y-5)x^2+(2-2y)x-(y+1)=0.$$ (The roots $x=1,-\frac13$ of the denominator cannot occur, since the numerator is non-zero there.) Step 2 - Discriminant. By (1), $$4(1-y)^2+4(3y-5)(y+1)=16\left(y^2-y-1\right)\ge0.$$ So $y\le\frac{1-\sqrt5}{2}$ or $y\ge\frac{1+\sqrt5}{2}$. (For $y=\frac53$ the equation is linear and still has a root $x=-2$.) Step 3 - Back to $t$. $\sin t=\frac y2$, so by (2) $$\sin t\le-\frac{\sqrt5-1}{4}=\sin\left(-\frac{\pi}{10}\right)\quad\text{or}\quad\sin t\ge\frac{\sqrt5+1}{4}=\sin\frac{3\pi}{10}.$$ Step 4 - Range. $\sin$ increases on $\left[-\frac\pi2,\frac\pi2\right]$, so $$t\in\left[-\frac{\pi}{2},-\frac{\pi}{10}\right]\cup\left[\frac{3\pi}{10},\frac{\pi}{2}\right].$$ Answer: $t\in\left[-\dfrac{\pi}{2},-\dfrac{\pi}{10}\right]\cup\left[\dfrac{3\pi}{10},\dfrac{\pi}{2}\right]$