[JEE Advanced 2014] for x in 0 pi the equation sin x 2 sin 2x sin 3x
For $x\in(0,\pi)$, the equation $\sin x+2\sin2x-\sin3x=3$ has
(a) infinitely many solutions
(b) three solutions
(c) one solution
(d) no solution
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (d) no solution
$\sin x-\sin3x=-2\cos2x\sin x$, so LHS $=2\sin x(1-\cos2x)+\,$… its maximum on $(0,\pi)$ is strictly less than $3$; hence no solution.
JEE Advanced 2014 · Trigonometry — verified solution by the Vidaara Team.
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