JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 4)

The number of values of $$\theta $$ $$ \in $$ (0, $$\pi $$) for which the system of linear equations

x + 3y + 7z = 0

$$-$$ x + 4y + 7z = 0

(sin3$$\theta $$)x + (cos2$$\theta $$)y + 2z = 0.

has a non-trival solution, is -
two
one
four
three

Explanation

$$\left| {\matrix{ 1 & 3 & 7 \cr { - 1} & 4 & 7 \cr {\sin 3\theta } & {\cos 2\theta } & 2 \cr } } \right| = 0$$

(8 $$-$$ 7 cos 2$$\theta $$) $$-$$ 3($$-$$2 $$-$$ 7 sin 3$$\theta $$)

      +7 ($$-$$ cos 2$$\theta $$ $$-$$ 4 sin 3$$\theta $$) = 0

14 $$-$$ 7 cos 2$$\theta $$ + 21 sin 3$$\theta $$ $$-$$ 7 cos 2$$\theta $$

      $$-$$ 28 sin 3$$\theta $$ = 0

14 $$-$$ 7 sin 3$$\theta $$ $$-$$ 14 cos 2$$\theta $$ = 0

14 $$-$$ 7 (3 sin $$\theta $$ $$-$$ 4 sin3$$\theta $$ ) $$-$$ 14 (1 $$-$$ 2 sin2 $$\theta $$) = 0

$$-$$ 21 sin $$\theta $$ + 28 sin3 $$\theta $$ + 28 sin2 $$\theta $$ = 0

7 sin $$\theta $$ [$$-$$ 3 + 4 sin2 $$\theta $$ + 4 sin $$\theta $$] = 0 sin$$\theta $$,

4 sin2 $$\theta $$ + 6 sin $$\theta $$ $$-$$ 2 sin $$\theta $$ $$-$$ 3 = 0

2 sin $$\theta $$(2 sin $$\theta $$ + 3) $$-$$ 1 (2 sin $$\theta $$ + 3) = 0

sin $$\theta $$ = $${{ - 3} \over 2}$$; sin$$\theta $$ = $${1 \over 2}$$

Hence, 2 solutions in (0, $$\pi $$)

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