Trignometric Equations MCQ Questions & Answers in Trigonometry | Maths

We have,
$$\eqalign{ & 4\cos x\left( {2 - 3\,{{\sin }^2}x} \right) + \left( {\cos 2x + 1} \right) = 0 \cr & \Rightarrow 4\cos x\left( {3\,{{\cos }^2}x - 1} \right) + 2\,{\cos ^2}x = 0 \cr & \Rightarrow 2\cos x\left( {6\,{{\cos }^2}x + 2} \right)\left( {2\cos x - 1} \right) = 0 \cr & \Rightarrow 2\cos x\left( {3\cos x + 2} \right)\left( {2\cos x - 1} \right) = 0 \cr & \Rightarrow {\text{either}}\,\,\cos x = 0{\text{ which gives }}x = \frac{\pi }{2} \cr} $$
$${\text{or }}\cos x = - \frac{2}{3}$$
Which gives no value of $$x$$ for which $$0 \leqslant x \leqslant \frac{\pi }{2}$$   or $$\cos x = \frac{1}{2},$$   which gives $$x = \frac{\pi }{3}$$
So, the required difference $$ = \frac{\pi }{2} - \frac{\pi }{3} = \frac{\pi }{6}$$

$$\eqalign{ & \sin \pi \left( {{x^2} + x} \right) = \sin \pi {x^2} \cr & \Rightarrow \,\,\pi \left( {{x^2} + x} \right) = n\pi + {\left( { - 1} \right)^n}\pi {x^2} \cr & \therefore \,\,{x^2} + x = 2m + {x^2} \cr & \Rightarrow \,\,x = 2m \in {\Bbb Z} \cr} $$
or, $${x^2} + x = p' - {x^2},$$    where $$p'$$ is an odd integer
$$ \Rightarrow \,\,2{x^2} + x - p' = 0.$$
For nonintegral solution, $$x = \frac{{ - 1 \pm \sqrt {1 + 8p'} }}{4} = \frac{{ - 1 + \sqrt {1 + 8p'} }}{4}$$        (taking the positive value)
$$x = \frac{{\sqrt p - 1}}{4},$$   where $$p$$ is an odd integer.

Here, $${x^3} + {\left( {x + 2} \right)^2} + 2\sin x = 4.$$     Clearly, $$x = 0$$  satisfies the equation.
$$\eqalign{ & {\text{If }}0 < x \leqslant \pi ,{x^3} + {\left( {x + 2} \right)^2} + 2\sin x > 4. \cr & {\text{If }}\pi < x \leqslant 2\pi ,{x^3} + {\left( {x + 2} \right)^2} + 2\sin x > 27 + 25 - 2. \cr} $$
So, $$x = 0$$  is the only solution.

The condition for solutions is $$ - \sqrt {{p^2} + {q^2}} \leqslant r \leqslant \sqrt {{p^2} + {q^2}} .$$

We have,
$$\eqalign{ & {\sin ^4}x - \left( {k + 2} \right){\sin ^2}x - \left( {k + 3} \right) = 0 \cr & \Rightarrow {\sin ^2}x = \frac{{\left( {k + 2} \right) \pm \sqrt {{{\left( {k + 2} \right)}^2} + 4\left( {k + 3} \right)} }}{2} \cr & = \frac{{\left( {k + 2} \right) \pm \left( {k + 4} \right)}}{2} \cr & \Rightarrow {\sin ^2}x = k + 3\,\left( {\because {{\sin }^2}x = - 1{\text{ is not possible}}} \right) \cr & {\text{Since }}0 \leqslant {\sin ^2}x \leqslant 1, \cr & \therefore 0 \leqslant k + 3 \leqslant 1 \cr & {\text{or }} - 3 \leqslant k \leqslant - 2 \cr} $$

$$\max \sin \theta = 1.$$    So, the equation can have a solution only when $$\sin x = 1,\sin y = 1$$
$$\eqalign{ & \Rightarrow \,\,x = \frac{\pi }{2},y = \frac{\pi }{2} \cr & \therefore \,\,x + y = \pi . \cr} $$

$$\eqalign{ & {\cos ^2}x = 2\cos x \cdot \left( {1 - 3{{\cos }^2}x} \right) \cr & \Rightarrow \,\,\cos x\left\{ {6{{\cos }^2}x + \cos x - 2} \right\} = 0 \cr} $$
$$\therefore \,\,\cos x = 0,\frac{1}{2}, - \frac{2}{3}.$$    The two smallest positive values of $$x$$ are $$\frac{\pi }{3}$$ and $$\frac{\pi }{2}.$$

$$\eqalign{ & {\max _{\theta \in R}}\left\{ {5\sin \,\theta + 3\,\sin \left( {\theta - \alpha } \right)} \right\} = 7 \cr & \Rightarrow {\max _{\theta \in R}}\left\{ {5\sin \,\theta + 3\,\sin \,\theta \cos \,\alpha - 3\cos \theta \,\sin \,\alpha } \right\} = 7 \cr & \Rightarrow {\max _{\theta \in R}}\left\{ {\sin \left( {5 + 3\,\cos \,\alpha } \right) + \cos \,\theta \left( { - 3\sin \,\alpha } \right)} \right\} = 7 \cr & {\text{As }}\sqrt {{a^2} + {b^2}} \leqslant a\,\sin \,x + b\,\cos \,x \leqslant \sqrt {{a^2} + {b^2}} \cr & \therefore \,\sqrt {{{\left( {5 + 3\,\cos \,\alpha } \right)}^2} + {{\left( { - 3\sin \,\alpha } \right)}^2}} = 7 \cr & \Rightarrow 34 + 30\cos \,\alpha = 49 \cr & \Rightarrow \cos \,\alpha = \frac{1}{2} \cr & \Rightarrow \alpha = 2n\pi \pm \frac{\pi }{3} \cr} $$

$$\cos \left( {x + \frac{\pi }{4}} \right) \geqslant \frac{1}{{\sqrt 2 }}.$$    The value scheme for this is shown below.

From the figure,
$$ - \frac{\pi }{4} \leqslant x + \frac{\pi }{4} \leqslant \frac{\pi }{4}$$    and in general, $$2n\pi - \frac{\pi }{4} \leqslant x + \frac{\pi }{4} \leqslant 2n\pi + \frac{\pi }{4}$$
$$\eqalign{ & \therefore \,\,2n\pi - \frac{\pi }{2} \leqslant x \leqslant 2n\pi . \cr & {\text{For, }}n = 0, - \frac{\pi }{2} \leqslant x \leqslant 0;\,{\text{for }}n = 1,\frac{{3\pi }}{2} \leqslant x \leqslant 2\pi . \cr} $$

Here, $$\left( {2\sin x - 1} \right)\left( {2\sin x - 3} \right) \leqslant 0.$$      But $$2\sin x - 3$$   is always negative.

$$\therefore \,\,2\sin x - 1 \geqslant 0,\,{\text{i}}{\text{.e}}{\text{.,}}\sin x \geqslant \frac{1}{2}.$$
∴ from the figure, $$\frac{\pi }{6} \leqslant x \leqslant \frac{{5\pi }}{6}.$$