Trignometric Equations MCQ Questions & Answers in Trigonometry | Maths

The given equation is $$\tan x + \sec x = 2\cos x;$$
$$\eqalign{ & \Rightarrow \sin x + 1 = 2{\cos ^2}x \cr & \Rightarrow \sin x + 1 = 2\left( {1 - {{\sin }^2}x} \right); \cr & \Rightarrow 2{\sin ^2}x + \sin x - 1 = 0; \cr & \Rightarrow \left( {2\sin x - 1} \right)\left( {\sin x + 1} \right) = 0 \cr & \Rightarrow \sin x = \frac{1}{2}, - 1.; \cr & \Rightarrow x = {30^ \circ },{150^ \circ },{270^ \circ }. \cr} $$

$$2\sin x + 1 \geqslant 0$$
$$ \Rightarrow \,\,\sin x \geqslant - \frac{1}{2}.$$   The value scheme for this is shown below.

From the figure,
$$\eqalign{ & \frac{{11\pi }}{6} \leqslant x \leqslant 2\pi \,\,{\text{or, }}0 \leqslant x \leqslant \frac{{7\pi }}{6}. \cr & \therefore \,\,x \in \left[ {0,\frac{{7\pi }}{6}} \right] \cup \left[ {\frac{{11\pi }}{6},2\pi } \right]. \cr} $$

$$\eqalign{ & \left( {\cos \frac{x}{4} - 2\sin x} \right)\sin x + \left( {1 + \sin \frac{x}{4} - 2\cos x} \right)\cos x = 0 \cr & \Rightarrow \left( {\sin x\cos \frac{x}{4} + \cos x\sin \frac{x}{4}} \right) + \cos x - 2\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0 \cr & \Rightarrow \sin \left( {x + \frac{x}{4}} \right) + \cos x - 2\left( 1 \right) = 0 \cr & \Rightarrow \sin \frac{{5x}}{4} + \cos x = 2 \cr & \Rightarrow \sin \frac{{5x}}{4} = \cos x = 1 \cr & \Rightarrow \sin \frac{{5x}}{4} = 1 \cr & \Rightarrow \frac{{5x}}{4} = 2n\pi + \frac{\pi }{2} \cr & \Rightarrow x = \frac{{8n\pi }}{5} + \frac{{2\pi }}{5}\,\,\& \,\,\cos x = 1 \cr & \Rightarrow x = 2m\pi \cr & {\text{Thus we have, }}\frac{{8n\pi }}{5} + \frac{{2\pi }}{5} = 2m\pi \cr & \Rightarrow m = \frac{{4n + 1}}{5} \cr} $$
$$\therefore n \in I,$$   so $$m$$ must be of the form $$m = 5k + 1$$
Hence, the solution of the equation is
$$x = 2\left( {5k + 1} \right)\pi ,k \in I$$

The equation holds if $$\left| {\cos x} \right| = 1{\text{ i}}{\text{.e}}{\text{., if }}x = n\pi ,n \in I$$
If $$\left| {\cos x} \right| \ne 1$$   then $${\sin ^2}x - \frac{3}{2}\sin x + \frac{1}{2} = 0$$
$$ \Rightarrow \sin x = 1{\text{ or }}\frac{1}{2}$$
$$\sin x \ne 1,$$   as in that case $$\cos x = 0$$
$$\eqalign{ & \therefore \sin x = \frac{1}{2} \cr & \Rightarrow x = n\,\pi + {\left( { - 1} \right)^n}\frac{\pi }{6} \cr} $$

$${\cos ^2}\theta = \frac{1}{6}\sin \theta \cdot \tan \theta \,\,\,{\text{or, }}6{\cos ^3}\theta + {\cos ^2}\theta - 1 = 0.$$
As $$\cos \theta = \frac{1}{2}$$   satisfies the equation (by trial),
$$\left( {2\cos \theta - 1} \right)\left( {3{{\cos }^2}\theta + 2\cos \theta + 1} \right) = 0$$
$$ \Rightarrow \cos \theta = \frac{1}{2}$$   (other values of $$\cos \theta $$  are imaginary).
So, $$\theta = 2n\pi \pm \frac{\pi }{3},n \in {\Bbb Z}.$$

We have $$\left| {\sqrt 3 \cos x - \sin x} \right| \leqslant \sqrt {3 + 1} = 2.$$      Thus, we must have $$\left| {\sqrt 3 \cos x - \sin x} \right| = 2$$
$$\eqalign{ & \therefore \,\,\left| {\cos \left( {x + \frac{\pi }{6}} \right)} \right| = 1 \cr & \therefore \,\,\cos \left( {x + \frac{\pi }{6}} \right) = 1, - 1 \cr & \therefore \,\,x + \frac{\pi }{6} = 0,2\pi ,4\pi ,6\pi ,.....,\pi ,3\pi ,5\pi ,..... \cr & \therefore \,\,x = \frac{{11\pi }}{6},\frac{{23\pi }}{6},\frac{{5\pi }}{6},\frac{{17\pi }}{6}. \cr} $$

$$\eqalign{ & {\text{Here, }}\sin \theta = - \frac{1}{2}\,{\text{and}}\tan \theta = \frac{1}{{\sqrt 3 }}. \cr & \sin \theta = - \frac{1}{2} \cr & \Rightarrow \,\,\theta = m\pi + {\left( { - 1} \right)^m} \cdot \left( { - \frac{\pi }{6}} \right)\,\,{\text{and}}\tan \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \,\,\theta = m\pi + \frac{\pi }{6}. \cr} $$
For common values, $$m$$ must be odd, i.e., $$m = 2n + 1.$$

Here, $$\left( {2\sec x + 1} \right)\left( {\sec x - 3} \right) = 0;$$      but $$\left| {\sec x} \right| \geqslant 1.$$   So, $$\sec x = 3,$$   which gives two values of $$\theta $$ in each of $$\left[ {0,2\pi } \right],\left( {2\pi ,4\pi } \right],\left( {4\pi ,6\pi } \right],$$     and one value in $$\left( {6\pi ,6\pi + \frac{{3\pi }}{2}} \right].$$

$$\eqalign{ & 2{\sin ^2}x + 5\sin x - 3 = 0 \cr & \Rightarrow \,\,\left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0 \cr & \Rightarrow \,\,\sin x = \frac{1}{2}\,\,\,{\text{and }}\sin x \ne - 3 \cr & \therefore \,\,{\text{In }}\left[ {0,3\pi } \right],x\,\,{\text{has 4 values}}{\text{.}} \cr} $$

$$\eqalign{ & 1 - \cos x = \left( {\sqrt 2 - 1} \right)\sin x \cr & \Rightarrow \,\,\sin \frac{x}{2}\left\{ {\sin \frac{x}{2} - \left( {\sqrt 2 - 1} \right)\cos \frac{x}{2}} \right\} = 0 \cr & \therefore \,\,\sin \frac{x}{2} = 0\,\,\,\,{\text{or, }}\tan \frac{x}{2} = \sqrt 2 - 1 = \tan \frac{{{{45}^ \circ }}}{2} \cr & \Rightarrow \,\,\frac{x}{2} = n\pi \,\,\,{\text{or, }}\frac{x}{2} = n\pi + \frac{\pi }{8} \cr & \therefore \,\,x = 2n\pi ,2n\pi + \frac{\pi }{4}. \cr} $$