Trignometric Equations MCQ Questions & Answers in Trigonometry | Maths

The maximum values of $$\sin x + \cos x$$   and $$1 + \sin 2x$$   are $$\sqrt 2 $$ and $$2$$ respectively. Also, $${\left( {\sqrt 2 } \right)^2} = 2.$$
∴ the equation can hold only when $$\sin x + \cos x = \sqrt 2 \,{\text{and }}1 + \sin 2x = 2.$$
Now, $$\sin x + \cos x = \sqrt 2 $$
$$\eqalign{ & \Rightarrow \,\,\cos \left( {x - \frac{\pi }{4}} \right) = 1 \cr & \Rightarrow \,\,x = 2n\pi + \frac{\pi }{4}. \cr & 1 + \sin 2x = 2 \cr & \Rightarrow \,\,\sin 2x = 1 \cr & \Rightarrow \,\,x = \frac{{n\pi }}{2} + {\left( { - 1} \right)^n} \cdot \frac{\pi }{4}. \cr} $$
The value in $$\left[ { - \pi ,\pi } \right]$$  satisfying both the equations is $$\frac{\pi }{4}\left( {{\text{when }}n = 0} \right).$$

Here, $$\sin x = 1 \pm \sqrt 2 ;$$    but $$\left| {\sin x} \right| \leqslant 1.$$   So, $$\sin x = 1 - \sqrt 2 ,\,$$   which gives two values of $$\theta $$ in each of $$\left[ {0,2\pi } \right],\left( {2\pi ,4\pi } \right],\left( {4\pi ,6\pi } \right],\,\,{\text{e}}{\text{.t}}{\text{.c}}{\text{.}}$$      Hence, the least value of $$n = 4.$$

Eliminating $$\theta ,r = 4\left( {1 + \frac{3}{r}} \right)$$
$$\eqalign{ & \Rightarrow \,\,\left( {r - 6} \right)\left( {r + 2} \right) = 0 \cr & \Rightarrow \,\,r = 6. \cr & \therefore \,\,\sin \theta = \frac{1}{2} \cr & \Rightarrow \,\,\theta = \frac{\pi }{6},\frac{{5\pi }}{6}.\,\,{\text{So, }}\left( {r,\theta } \right) = \left( {6,\frac{\pi }{6}} \right),\left( {6,\frac{{5\pi }}{6}} \right). \cr} $$

Clearly, $$x \geqslant 4$$  (Since $${\sqrt {x - 4} }$$   is real) so that $${\sqrt x }\,$$ is also real.
Again, if $$\cos \left( {\pi \sqrt x } \right) < 1$$    then
$$\cos \left( {\pi \sqrt {x - 4} } \right) > 1$$    and if $$\cos \left( {\pi \sqrt x } \right) > 1,$$    then $$\cos \left( {\pi \sqrt {x - 4} } \right) < 1$$    (since this product $$= 1$$  ).
But both of these are not possible (since $$\cos \theta$$  cannot be greater than 1).
$$\eqalign{ & \therefore \cos \left( {\pi \sqrt {x - 4} } \right) = 1{\text{ and }}\cos \left( {\pi \sqrt x } \right) = 1 \cr & \therefore x - 4 = 0{\text{ or }}x = 0 \cr} $$
But $$x = 0$$  is not possible,
$$\therefore x = 4$$   is the only solution.

$$\eqalign{ & {\text{Here,}}\,\,p\,\sin x = \frac{\pi }{2} \pm p\cos x\,\,\,{\text{or,}}\,\,\sin x \mp \,\cos x = \frac{\pi }{{2\,p}} \cr & {\text{or,}}\,\,{\text{sin}}\left( {x \mp \frac{\pi }{4}} \right) = \frac{\pi }{{2\sqrt {2\,p} }} \cr & \Rightarrow \,\left| {\frac{\pi }{{2\sqrt {2\,p} }}} \right| \leqslant 1 \cr} $$
∴ for positive $$p,p \geqslant \frac{\pi }{{2\sqrt 2 }}.\,{\text{But}}\,{\text{1}}\,{\text{ < }}\frac{\pi }{{2\sqrt 2 }} < 2.$$

$$\eqalign{ & 2\,{\sin ^2}\theta - \cos 2\theta = 0 \cr & \Rightarrow \,\,1 - 2\cos 2\theta = 0 \cr & \Rightarrow \,\,\cos 2\theta = \frac{1}{2} \cr & \Rightarrow \,\,2\theta = \frac{\pi }{3},\frac{{5\pi }}{3},\frac{{7\pi }}{3},\frac{{11\pi }}{3} \cr & \Rightarrow \,\,\theta = \frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6},\frac{{11\pi }}{6}\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{where }}\theta \in \left[ {0,2\pi } \right] \cr & {\text{Also }}2\,{\cos ^2}\theta - 3\sin \theta = 0 \cr & \Rightarrow \,\,2\,{\sin ^2}\theta + 3\sin \theta - 2 = 0 \cr & \Rightarrow \,\,\left( {2\sin \theta - 1} \right)\left( {\sin \theta + 2} \right) = 0 \cr & \Rightarrow \,\,\sin \theta = \frac{1}{2}\,\,\,\left[ {\because \,\,\sin \theta \ne - 2} \right] \cr & \Rightarrow \,\,\theta = \frac{\pi }{6},\frac{{5\pi }}{6}\,\,\,\,.....\left( 2 \right),\,\,{\text{where}}\,\,\theta \in \left[ {0,2\pi } \right] \cr} $$
Combining (1) and (2), we get $$\theta = \frac{\pi }{6},\frac{{5\pi }}{6}$$
∴ Two solutions are there.

$$\eqalign{ & 13\cos \theta + 12\sin \theta = 5 \cr & {\text{or, }}\frac{{13}}{{\sqrt {{{13}^2} + {{12}^2}} }}\cos \theta + \frac{{12}}{{\sqrt {{{13}^2} + {{12}^2}} }}\sin \theta = \frac{5}{{\sqrt {{{13}^2} + {{12}^2}} }} \cr & {\text{or, }}\cos \left( {\theta - \alpha } \right) = \frac{5}{{\sqrt {313} }},\,{\text{where }}\cos \alpha = \frac{{13}}{{\sqrt {313} }} \cr & \therefore \,\,\theta = 2n\pi \pm {\cos ^{ - 1}}\frac{5}{{\sqrt {313} }} + \alpha = 2n\pi \pm {\cos ^{ - 1}}\frac{5}{{\sqrt {313} }} + {\cos ^{ - 1}}\frac{{13}}{{\sqrt {313} }}. \cr} $$
As $${\cos^{ - 1}}\frac{5}{{\sqrt {313} }} > {\cos ^{ - 1}}\frac{{13}}{{\sqrt {313} }},$$      we get $$\theta \in \left[ {0,2\pi } \right],$$   when $$n = 0$$  (one value, taking positive sign) and when $$n = 1$$  (one value, taking negative sign).

$$\frac{{{2^{\sin x}} + {2^{\cos x}}}}{2} \geqslant \sqrt {{2^{\sin x}} \cdot {2^{\cos x}}} $$      (using AM $$ \geqslant $$ GM where the equality holds when $${2^{\sin x}} = {2^{\cos x}}$$  )
$$\therefore \,\,{2^{\sin x}} + {2^{\cos x}} \geqslant 2 \cdot \sqrt {{2^{\sin x + \cos x}}} .$$
But the minimum value of $$\sin x + \cos x\,\,{\text{is }} - \sqrt 2 .$$
$$\therefore \,\,{2^{\sin x}} + {2^{\cos x}} \geqslant 2 \cdot \sqrt {{2^{ - \sqrt 2 }}} = {2^{1 - \frac{1}{{\sqrt 2 }}}}.$$
The equation holds when the equality holds, i.e., $${2^{\sin x}} = {2^{\cos x}}.$$
$$\therefore \,\,\sin x = \cos x\,\,\,{\text{or, }}\tan x = 1$$

The condition is $$\left| {\frac{{k + 1}}{{\sqrt {{k^2} + 9} }}} \right| \leqslant 1,\,{\text{i}}{\text{.e}}{\text{., }}{\left( {k + 1} \right)^2} \leqslant {k^2} + 9.\,\,\,\,\therefore \,\,k \leqslant 4.$$

$$\eqalign{ & {\text{Given that}}:\sin \left( {\pi \cos x} \right) = \cos \left( {\pi \sin x} \right) \cr & {\text{So}},\,\,\cos \left( {\frac{\pi }{2} - \pi \cos x} \right) = \cos \left( {\pi \sin x} \right) \cr & \Rightarrow \frac{\pi }{2} - \pi \cos x = \pi \sin x \cr & \Rightarrow \sin x + \cos x = \frac{1}{2} \cr} $$
Squaring both sides, we get
$$\eqalign{ & {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x = \frac{1}{4} \cr & \Rightarrow \sin 2x = \frac{1}{4} - 1 = - \frac{3}{4} \cr} $$