Trignometric Equations MCQ Questions & Answers in Trigonometry | Maths

$$ - 1 \leqslant \tan \theta \leqslant 1.$$    The value scheme for this is shown below.

From the figure,
$$\eqalign{ & - \frac{\pi }{4} \leqslant x \leqslant \frac{\pi }{4}\,\,\,{\text{or, }} - \pi \leqslant x \leqslant - \frac{{3\pi }}{4}\,\,\,{\text{or, }}\frac{{3\pi }}{4} \leqslant x \leqslant \pi . \cr & \therefore \,\,x \in \left[ { - \pi , - \frac{{3\pi }}{4}} \right] \cup \left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right] \cup \left[ {\frac{{3\pi }}{4},\pi } \right]. \cr} $$

$$\eqalign{ & {16^{{{\sin }^2}x}} + \frac{{16}}{{{{16}^{{{\sin }^2}x}}}} = 10\,\,\,{\text{or, }}{y^2} - 10y + 16 = 0,\,\,{\text{where }}y = {16^{{{\sin }^2}x}}. \cr & \therefore \,\,y = 2,8 \cr & \Rightarrow \,\,\,{16^{{{\sin }^2}x}} = {2^1},{2^3} \cr & \Rightarrow \,\,4{\sin ^2}x = 1,3 \cr & \therefore \,\,\,\sin x = \frac{1}{2}, - \frac{1}{2},\frac{{\sqrt 3 }}{2}, - \frac{{\sqrt 3 }}{2}. \cr} $$
Each value will give two values of $$x$$ in $$\left[ {0,2\pi } \right].$$

$$\eqalign{ & \cos 7\theta = \cos\theta - \sin4\theta \cr & \Rightarrow \sin 4\theta = \cos \theta - \cos 7\theta \cr & \Rightarrow \sin 4\theta = 2\sin 4\theta \sin 3\theta \cr & \Rightarrow \sin 4\theta \left( {1 - 2\sin 3\theta } \right) = 0 \cr & \therefore \sin 4\theta = 0{\text{ or }}\,\sin 3\theta = \frac{1}{2} \cr & \Rightarrow 4\theta = n\pi {\text{ or }}\,3\theta = n\pi + {\left( { - 1} \right)^n}\frac{\pi }{6} \cr & \Rightarrow \theta = \frac{{n\pi }}{4}{\text{ or }}\frac{{n\pi }}{3} + {\left( { - 1} \right)^n}\frac{\pi }{{18}} \cr} $$

The given equation is
$$\eqalign{ & 2\,{\cos ^2}\left( {\frac{x}{2}} \right){\sin ^2}x = {x^2} + \frac{1}{{{x^2}}}\,\,{\text{where }}0 < x \leqslant \frac{\pi }{2} \cr & {\text{L.H.S.}} = 2\,{\cos ^2}\frac{x}{2}{\sin ^2}x = \left( {1 + \cos x} \right){\sin ^2}x \cr & \because \,\,1 + \cos x < 2\,\,{\text{and }}{\sin ^2}x \leqslant 1\,\,{\text{for }}0 < x \leqslant \frac{\pi }{2} \cr & \therefore \,\,\left( {1 + \cos x} \right){\sin ^2}x < 2 \cr & {\text{And R}}{\text{.H}}{\text{.S}}{\text{. }} = {x^2} + \frac{1}{{{x^2}}} \geqslant 2 \cr & \therefore \,\,{\text{For }}0 < x \leqslant \frac{\pi }{2}. \cr} $$
given equation is not possible for any real value of $$x.$$

$$\eqalign{ & f\left( {\frac{\pi }{{10}}} \right) = \sin {18^ \circ } + \cos {18^ \circ } = \sqrt 2 \sin \left( {{{45}^ \circ } + {{18}^ \circ }} \right) = \sqrt 2 \sin {63^ \circ }. \cr & {\text{As }}\sin {63^ \circ } > \sin {45^ \circ } = \frac{1}{{\sqrt 2 }}\,{\text{and}}\sin {63^ \circ } < 1,\,{\text{we get }}1 < f\left( {\frac{\pi }{{10}}} \right) < \sqrt 2 . \cr} $$
$$\therefore \,\,\left[ {f\left( {\frac{\pi }{{10}}} \right)} \right] = 1.$$    So, the equation is $$\sin x + \cos x = 1.$$
$$\therefore \,\,\cos\left( {x - \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}\,\,{\text{or, }}x - \frac{\pi }{4} = 2n\pi \pm \frac{\pi }{4}.$$

$$\eqalign{ & {\text{Since, }}{x^2} + {x^{ - 2}} = {\left( {x - {x^{ - 1}}} \right)^2} + 2 \leqslant 2 \cr & {\text{and }}2\,{\cos ^2}\frac{x}{2}{\sin ^2}x \leqslant 2, \cr} $$
$$\therefore $$ the given equation is valid only if
$$2\,{\cos ^2}\frac{x}{2}{\sin ^2}x = 2$$
$$ \Leftrightarrow \cos \frac{x}{2} = \operatorname{cosec} x = 1,$$     which cannot be true.

$$\eqalign{ & {\tan ^2}x = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} = {\left( {\sqrt 2 - 1} \right)^2} = {\tan ^2}\frac{\pi }{8} \cr & \Rightarrow \,\,\tan x = \tan \left( { \pm \frac{\pi }{8}} \right). \cr} $$