Trignometric Equations MCQ Questions & Answers in Trigonometry | Maths

$$2\sin \theta = {\left( {{r^2} - 1} \right)^2} + 2 \geqslant 2.\,{\text{But}}\max \,\sin\theta = 1.$$
∴ the equation implies $$\sin\theta = 1,{r^2} = 1.$$
$$\eqalign{ & \sin\theta = 1 \cr & \Rightarrow \,\,\theta = \frac{\pi }{2},\frac{{5\pi }}{2},\frac{{9\pi }}{2}\left( {{\text{from the question}}} \right). \cr & {r^2} = 1 \cr & \Rightarrow \,\,r = 1, - 1. \cr} $$
∴ the number of values of the pair $$\left( {r,\theta } \right) = 2 \times 3 = 6.$$

$$\eqalign{ & 2\,{\cot ^2}\theta + 2\sqrt 3 \cot \theta + 4\,{\text{cosec}}\,\theta + 8 = 0 \cr & \Rightarrow {\cot ^2}\theta + {\operatorname{cosec} ^2}\theta - 1 + 2\sqrt 3 \cot \theta + 4\operatorname{cosec} \theta + 8 = 0\left( {\because {{\cot }^2}\theta = {{\operatorname{cosec} }^2}\theta - 1} \right) \cr & \Rightarrow \left( {{{\cot }^2}\theta + 2\sqrt 3 \cot \theta + 3} \right) + \left( {{{\operatorname{cosec} }^2}\theta + 4\operatorname{cosec} \theta + 4} \right) = 0 \cr & \Rightarrow {\left( {\cot \theta + \sqrt 3 } \right)^2} + {\left( {\operatorname{cosec} \theta + 2} \right)^2} = 0 \cr & \Rightarrow \cot \theta = - \sqrt 3 \,{\text{and }}\,\operatorname{cosec} \theta = - 2 \cr} $$
Principal value of $$\theta$$ satisfying both the equation is $$\theta = 2\pi - \frac{\pi }{6} = \frac{{11\pi }}{6}$$
$$\therefore \,$$ General solution is $$\theta = 2n\pi + \frac{{11\pi }}{6},n \in I$$

$$\eqalign{ & \max \cos \theta = 1.\,\,{\text{So, }}\cos x \cdot \sin y = 1 \cr & \Rightarrow \,\,\cos x = 1,\sin y = 1\,\,\,{\text{or, }}\cos x = - 1,\sin y = - 1. \cr & \cos x = 1,\sin y = 1 \cr & \Rightarrow \,\,x = 0,2\pi \,\,{\text{and }}y = \frac{\pi }{2},\frac{{5\pi }}{2}\left( {{\text{from the question}}} \right). \cr & \cos x = - 1,\sin y = - 1 \cr & \Rightarrow \,\,x = \pi ,3\pi \,\,{\text{and }}y = \frac{{3\pi }}{2}\left( {{\text{from the question}}} \right). \cr} $$
∴ the required number of ordered pair $$ = 2 \times 2 + 2 \times 1 = 6.$$

$$\eqalign{ & \because \,\,8\cos x.\left( {{{\cos }^2}\frac{\pi }{6} - {{\sin }^2}x - \frac{1}{2}} \right) = 1 \cr & \Rightarrow \,\,8\cos x\left( {\frac{3}{4} - \frac{1}{2} - {{\sin }^2}x} \right) = 1 \cr & \Rightarrow \,\,8\cos x\left( {\frac{1}{4} - \left( {1 - {{\cos }^2}x} \right)} \right) = 1 \cr & \Rightarrow \,\,8\cos x\left( {\frac{1}{4} - 1 + {{\cos }^2}x} \right) = 1 \cr & \Rightarrow \,\,8\cos x\left( {{{\cos }^2}x - \frac{3}{4}} \right) = 1 \cr & \Rightarrow \,\,8\left( {\frac{{4{{\cos }^3}x - 3\cos x}}{4}} \right) = 1 \cr & \Rightarrow \,\,2\left( {4{{\cos }^3}x - 3\cos x} \right) = 1 \cr & \Rightarrow \,\,2\cos 3x = 1 \cr & \Rightarrow \,\,\cos 3x = \frac{1}{2} \cr & \therefore \,\,3x = 2n\pi \pm \frac{\pi }{3},n \in 1 \cr & \Rightarrow \,\,x = \frac{{2n\pi }}{3} \pm \frac{\pi }{9} \cr & {\text{In }}x \in \left[ {0,\pi } \right]:x = \frac{\pi }{9},\frac{{2\pi }}{3} + \frac{\pi }{9},\frac{{2\pi }}{3} - \frac{\pi }{9},{\text{only}} \cr} $$
Sum of all the solutions of the equation
$$\eqalign{ & = \left( {\frac{1}{9} + \frac{2}{3} + \frac{1}{9} + \frac{2}{3} - \frac{1}{9}} \right)\pi \cr & = \frac{{13}}{9}\pi \cr} $$

The given eqn is, $$\tan x + \sec x = 2\cos x$$
$$\eqalign{ & \Rightarrow \,\,\frac{{\sin x}}{{\cos x}} + \frac{1}{{\cos x}} = 2\cos x \cr & \Rightarrow \,\,\sin x + 1 = 2{\cos ^2}x \cr & \Rightarrow \,\,2\,{\sin ^2}x + \sin x - 1 = 0 \cr & \Rightarrow \,\,\left( {2\sin x - 1} \right)\left( {\sin x + 1} \right) = 0 \cr & \Rightarrow \,\,\sin x = \frac{1}{2}, - 1 \cr & \Rightarrow \,\,x = \frac{\pi }{6},\frac{{5\pi }}{6},\frac{{3\pi }}{2} \in \left[ {0,2\pi } \right] \cr} $$
But for $$x = \frac{{3\pi }}{2},$$   given eqn. is not defined,
∴ only 2 solutions.

$$\eqalign{ & \tan 2\theta \cdot \tan \theta = 1 \cr & \Rightarrow \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} \cdot \tan \theta = 1 \cr & \Rightarrow 2\,{\tan ^2}\theta = 1 - {\tan ^2}\theta \cr & \Rightarrow 3\,{\tan ^2}\theta = 1 \cr & \Rightarrow {\tan ^2}\theta = \frac{1}{3} = {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & \Rightarrow {\tan ^2}\theta = {\tan ^2}\left( {{{30}^ \circ }} \right){\tan ^2}\left( {\frac{\pi }{6}} \right)\left[ {\because \theta = n\pi \pm \frac{\pi }{6}} \right]; \cr & \therefore \theta = \frac{\pi }{6} \cr} $$

As $$\max \cos \theta = 1,2\cos x + \cos 2\lambda x = 3$$        is possible only when $$\cos x = 1\,{\text{and}}\cos 2\lambda x = 1,\,{\text{i}}{\text{.e}}{\text{.,}}\cos x = 1\,{\text{and}}\sin \lambda x = 0.$$
Clearly, if $$\lambda $$ is rational, say $$\frac{p}{q},$$  then $$x = 2q\pi ,q \in {\Bbb Z},$$    satisfies both the equations. Therefore, for exactly one solution, $$x = 0,\,\,\,\lambda $$   should be irrational.

Here, $$6{\cos ^2}x - 10\cos x + 4 = 0\,\,\,\,{\text{or, }}\left( {3\cos x - 2} \right)\left( {\cos x - 1} \right) = 0$$
$$\eqalign{ & \therefore \,\,\cos x = \frac{2}{3},1 \cr & \Rightarrow \,\,x = 2n\pi \pm {\cos ^{ - 1}}\frac{2}{3},2n\pi . \cr} $$
From $$x = 2n\pi $$   we get one solution each for $$n = 0,1,2.$$
From $$x = 2n\pi \pm {\cos ^{ - 1}}\frac{2}{3}$$    we get the following solutions:
when $$n = 0,$$  one solution; when $$n = 1,$$  two solutions; when $$n = 2,$$  two solutions.

$$\eqalign{ & 1 + \sin \theta = \sqrt 3 \cos \theta \,\,\,{\text{or, }}\sqrt 3 \cos \theta - \sin \theta = 1 \cr & {\text{or, }}\cos \left( {\theta + \frac{\pi }{6}} \right) = \frac{1}{2}\,\,{\text{or, }}\theta + \frac{\pi }{6} = 2n\pi \pm \frac{\pi }{3}\,\,{\text{or, }}\theta = 2n\pi + \frac{\pi }{6},2n\pi - \frac{\pi }{2}. \cr} $$
∴ when $$\theta = 2n\pi + \frac{\pi }{6},$$   there are solutions for $$n = 0,1$$  and
when $$\theta = 2n\pi - \frac{\pi }{2},$$   for $$n = 1.$$  But this value does not satisfy the given equation.

Here, $$\cos x < \frac{{\sqrt 3 }}{2}.$$   The value scheme for this is shown below.

From the figure,
$$ - \pi \leqslant x < - \frac{\pi }{6}\,\,{\text{or, }}\frac{\pi }{6} < x \leqslant \pi .$$
$$\therefore \,\,x \in \left[ { - \pi , - \frac{\pi }{6}} \right) \cup \left( {\frac{\pi }{6},\pi } \right].$$