JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Trigonometric Equations

Rewrite the equation using ${\cos }^{2}x=1-{\sin }^{2}x$:



Using the quadratic formula:


Since $-1\le \sin x\le 1$, only $\sin x=\frac{-1+\sqrt{5}}{2}\approx 0.618$ is valid (the other root is negative and less than $-1$).

For $\sin x=\frac{-1+\sqrt{5}}{2}$ in $(0,10)$:

• In each period $[0,2\pi ]$, there are 2 solutions (one in $(0,\pi )$ and one in $(\pi ,2\pi )$ if the value is positive, but actually only in $(0,\pi )$ for positive sine).
• Actually, for $0<\sin x<1$, there are 2 solutions per period: one in $(0,\pi /2)$ and one in $(\pi /2,\pi )$.

The interval $(0,10)$ contains approximately $\frac{10}{2\pi }\approx 1.59$ periods, so about 3 complete periods.

In 3 periods: $3\times 2=6$ solutions. But checking more carefully with the interval length, we get 4 solutions in $(0,10)$.