Trignometric Equations MCQ Questions & Answers in Trigonometry | Maths

The given equation is $$2\,{\sin ^2}\theta - 3\sin \theta - 2 = 0$$
$$\eqalign{ & \Rightarrow \,\,\left( {2\sin \theta + 1} \right)\left( {\sin \theta - 2} \right) = 0 \cr & \Rightarrow \,\,\sin \theta = - \frac{1}{2}\,\,\,\left[ {\because \,\,\sin \theta - 2 = 0\,{\text{is not possible}}} \right] \cr & \Rightarrow \,\,\sin \theta = \sin \left( { - \frac{\pi }{6}} \right) = \sin \left( {\frac{{7\pi }}{6}} \right) \cr & \Rightarrow \,\,\theta = n\pi + {\left( { - 1} \right)^n}\left( { - \frac{\pi }{6}} \right) \cr & = n\pi + {\left( { - 1} \right)^n}\frac{{7\pi }}{6} \cr & \Rightarrow \,\,{\text{Thus,}}\,\theta = n\pi + {\left( { - 1} \right)^n}\frac{{7\pi }}{6} \cr} $$

$$\eqalign{ & \frac{{\sin x + i\cos x}}{{1 + i}} = \frac{{\left( {1 - i} \right)\left( {\sin x + i\cos x} \right)}}{{\left( {1 + i} \right)\left( {1 - i} \right)}} \cr & \frac{{\sin x + i\cos x}}{{1 + i}} = \frac{{\sin x + \cos x + i\left( {\cos x - \sin x} \right)}}{2},\,\,{\text{which is purely imaginary}} \cr & \Rightarrow \,\,\sin x + \cos x = 0 \cr & \Rightarrow \,\,\tan x = - 1 \cr & \therefore \,\,x = n\pi - \frac{\pi }{4}. \cr} $$

$$\eqalign{ & {2^{\frac{1}{{1 - \left| {\cos x} \right|}}}} = {2^2} \cr & \Rightarrow \,\,\frac{1}{{1 - \left| {\cos x} \right|}} = 2 \cr & \Rightarrow \,\,\left| {\cos x} \right| = \frac{1}{2} \cr & \Rightarrow \,\,\cos x = \pm \frac{1}{2}. \cr & \therefore \,\,x = 2n\pi \pm \frac{\pi }{3},2n\pi \pm \left( {\pi - \frac{\pi }{3}} \right) = 2n\pi \pm \frac{\pi }{3},\left( {2n \pm 1} \right)\pi \mp \frac{\pi }{3} = n\pi \pm \frac{\pi }{3}. \cr} $$

$$\eqalign{ & \sin x \cdot \cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right) > 0\,\,\,{\text{or, }}\frac{1}{2}\sin 2x \cdot \cos 2x > 0\,\,{\text{or, }}\sin 4x > 0 \cr & \therefore \,\,0 < 4x < \pi . \cr} $$

$$\eqalign{ & 2co{s^2}\theta + 3\sin \theta = 0 \cr & \Rightarrow \,\,\left( {2\sin \theta + 1} \right)\left( {\sin \theta - 2} \right) = 0 \cr & \Rightarrow \,\,\sin \theta = - \frac{1}{2}\,\,{\text{or sin}}\theta = 2 \to {\text{Not possible}} \cr} $$

The required sum of all solutions in $$\left[ { - 2\pi ,2\pi } \right]{\text{is}}$$
$$\eqalign{ & {\text{ = }}\left( {\pi + \frac{\pi }{6}} \right) + \left( {2\pi - \frac{\pi }{6}} \right) + \left( { - \frac{\pi }{6}} \right) + \left( { - \pi + \frac{\pi }{6}} \right) \cr & = 2\pi \cr} $$

$$\eqalign{ & {\text{If}}\,\cos x \geqslant 0,\,{\text{i}}{\text{.e}}{\text{., }}x \in \left[ {0,\frac{\pi }{2}} \right] \cup \left[ {\frac{{3\pi }}{2},\frac{{5\pi }}{2}} \right] \cup \left[ {\frac{{7\pi }}{2},4\pi } \right]{\text{then }}\cos x = \sin x. \cr & \therefore \,\,\tan x = 1\,\,\,{\text{or, }}x = n\pi + \frac{\pi }{4} = \frac{\pi }{4},\frac{{5\pi }}{4},\frac{{9\pi }}{4},\frac{{13\pi }}{4}. \cr} $$
∴ if $$\cos x \geqslant 0,$$   the possible values of $$x$$ are $$\frac{\pi }{4},\frac{{9\pi }}{4}.$$
$$\eqalign{ & {\text{If }}\cos x < 0,\,{\text{i}}{\text{.e}}{\text{., }}x \in \left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right) \cup \left( {\frac{{5\pi }}{2},\frac{{7\pi }}{2}} \right)\,{\text{then }} - \cos x = \sin x. \cr & \therefore \,\,\tan x = - 1\,\,\,\,{\text{or, }}x = n\pi - \frac{\pi }{4} = \frac{{3\pi }}{4},\frac{{7\pi }}{4},\frac{{11\pi }}{4},\frac{{15\pi }}{4}. \cr} $$
∴ if $$\cos x < 0$$   the possible values of $$x$$ are $$\frac{{3\pi }}{4},\frac{{11\pi }}{4}.$$

$$\eqalign{ & {\text{Given}}:2\sin \theta = x + \frac{1}{x} \cr & {\text{we know that}}\,\, - 1 \leqslant \sin \theta < 1, - 2 \leqslant 2\sin \theta < 2 \cr & {\text{So}},\,\, - 2 \leqslant x + \frac{1}{x} < 2 \cr} $$
Thus, the given equation is valid only if $$x = \pm 1$$

Here, $$3\,{\sin ^2}\theta = \cos 2\phi $$    and $$3\sin \theta \cdot \cos \theta = \sin 2\phi .$$     Squaring and adding, $$9\,{\sin^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1,\,{\text{i}}{\text{.e}}{\text{.,}}\,\,\sin \theta = \frac{1}{3}\,{\text{and }}\cos \theta = \frac{{2\sqrt 2 }}{3}.$$
$$\eqalign{ & \therefore \,\,\cos2\phi = 3 \cdot \frac{1}{9} = \frac{1}{3}\,\,{\text{and }}\sin 2\phi = \frac{{2\sqrt 2 }}{3}. \cr & \therefore \,\,\cos \left( {\theta + 2\phi } \right) = \cos \theta \cdot \cos 2\phi - \sin \theta \cdot \sin 2\phi \cr & \cos \left( {\theta + 2\phi } \right) = \frac{{2\sqrt 2 }}{3} \cdot \frac{1}{3} - \frac{1}{3} \cdot \frac{{2\sqrt 2 }}{3} = 0\,\,{\text{and }}\theta + 2\phi < \frac{{3\pi }}{2}. \cr} $$

Here, $$\frac{1}{2}\sin \left( {2{e^x}} \right) = \frac{1}{4}\left( {{2^x} + {2^{ - x}}} \right) \geqslant \frac{1}{4} \cdot 2 = \frac{1}{2};\,\,{\text{so,}}\sin \left( {2{e^x}} \right) \geqslant 1.$$
But $$\sin\left( {2{e^x}} \right) > 1.$$   Therefore, $$\sin\left( {2{e^x}} \right) = 1.$$
But equality can hold when $${2^x} = {2^{ - x}} = 1,\,{\text{i}}{\text{.e}}{\text{., }}x = 0.$$     Then $$\sin\left( {2 \cdot {e^0}} \right) = 1,$$   which is not true.

$$\eqalign{ & {\sin^2}3x + {\sin ^4}x = 0\,\,\,{\text{or, }}{\sin ^2}x\left\{ {{{\left( {3 - 4\,{{\sin }^2}x} \right)}^2} + {{\sin }^2}x} \right\} = 0 \cr & \therefore \,\,\sin x = 0 \cr & \Rightarrow \,\,x = n\pi . \cr} $$