IIT-JEE 2010 · Paper 1 · Mathematics Q55

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MathematicsInteger+3Question 55 of 84

The number of all possible values of $\theta$ , where $0<\theta<\pi$ , for which the system of equations
$(y+z)\cos3\theta=(xyz)\sin3\theta$
$x\sin3\theta=\frac{2\cos3\theta}y+\frac{2\sin3\theta}z$
$(xyz)\sin3\theta=(y+2z)\cos3\theta+y\sin3\theta$
have a solution $(x_0,y_0,z_0)$ with $y_0z_0\ne0$ , is

Your answer

Answer and worked solution

Official answer: 3

Write $s=\sin3\theta$ , $c=\cos3\theta$ .

Eliminate $xyz$ . (1) − (3): $(y+z)c=(y+2z)c+ys$ , so $ys+zc=0$ .
Multiply (2) by $yz$ : $xyz\,s=2zc+2ys$ ; with (1): $(y+z)c=2zc+2ys$ , i.e. $yc-zc-2ys=0$ . Using $ys=-zc$ this becomes $(y+z)c=0$ .

Cases.

$c=0$ : then $ys=0$ with $s=\pm1$ , so $y=0$ , which is excluded.
$c\ne0$ : $z=-y$ , and $ys-yc=0$ gives $\tan3\theta=1$ .

Solutions. $3\theta\in(0,3\pi)$ with $\tan3\theta=1$ : $3\theta=\frac\pi4,\frac{5\pi}4,\frac{9\pi}4$ . Each gives a solution ( $x=0$ , $z=-y\ne0$ ). Three values of θ.

Answer: 3

Topics:Trigonometric equations Systems of linear equations

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IIT-JEE 2010 · Paper 1, Mathematics question 55