Question

sin x + 2 sin 2x = 3 + sin 3x, 0 $\le$ x $\le 2\pi$ has

• A. 2 solutions in I quadrant
• B. One solution in II quadrant
• C. no solution in any quadrant
• D. one solution in each quadrant

Answer 1

The correct option is C no solution in any quadrant

From the given relation we have $sinx–sin3x+2sin2x=3\Rightarrow 2sinxcos2x–2sin2x+3=0$
$\Rightarrow (sinx+cos2x{)}^2+(sin2x–1{)}^2+3=si{n}^{2}x+co{s}^{2}2x+si{n}^{2}2x+1$
$\Rightarrow (sinx+cos2x{)}^2+(sin2x–1{)}^2+co{s}^{2}x=0$ which is possible only if
sin x + cos 2x = 0, sin 2x = 1 and cos x = 0 which is not possible for any value of x.