Trignometric Equations MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & {\text{Here, }}\left( {2\cos x + 3} \right)\left( {\cos x - 1} \right) = 0;\,{\text{but }}\left| {\cos x} \right| \leqslant 1. \cr & {\text{So, }}\cos x = 1 \cr & \Rightarrow \,\,x = 2n\pi \cr & \therefore \,\,x = 0,2\pi . \cr} $$
None of these values satisfies the given equation.

$$\eqalign{ & {a_1} + {a_2}\left( {2{{\cos }^2}x - 1} \right) + {a_3}\left( {1 - {{\cos }^2}x} \right) = 1 \cr & {\text{or, }}\left( {2{a_2} - {a_3}} \right){\cos ^2}x + \left( {{a_1} - {a_2} + {a_3} - 1} \right) = 0. \cr} $$
This can hold for all $$x$$ if $$2{a_2} - {a_3} = 0\,\,{\text{and }}{a_1} - {a_2} + {a_3} - 1 = 0.$$
As there are two equations in three unknowns, the number of solutions is infinite.

$$\left| {\frac{5}{{\sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}} }}} \right| \leqslant 1$$     is not true. So, there is no solution.

$$\eqalign{ & {\text{Let, }}\sqrt 3 + 1 = r\cos \alpha ,\,\,{\text{and }}\sqrt 3 - 1 = r\sin \alpha \cr & \therefore {r^2} = {\left( {\sqrt 3 + 1} \right)^2} + {\left( {\sqrt 3 - 1} \right)^2} = 8{\text{ i}}{\text{.e}}{\text{.,}}\,\alpha = \frac{\pi }{{12}} \cr & {\text{From the equation}},\,\,r\cos \left( {\theta - \alpha } \right) = 2 \cr & \Rightarrow \cos \left( {\theta - \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 2 }} = \cos \left( {\frac{\pi }{4}} \right) \cr & \therefore \theta = 2n\pi \pm \frac{\pi }{4} + \frac{\pi }{{12}} \cr} $$

$$\eqalign{ & {\text{We have, }}\,{\text{1}} - \cos 2x + 1 - {\cos ^2}2x = 2 \cr & {\text{or, }}\,\cos 2x\left( {\cos 2x + 1} \right) = 0 \cr & \therefore \cos 2x = 0, - 1, \cr & \therefore 2x = \left( {n + \frac{1}{2}} \right)\pi \,{\text{or}}\,\left( {2n + 1} \right)\pi \cr & \Rightarrow x = \left( {2n + 1} \right)\frac{\pi }{4}\,{\text{or}}\,\left( {2n + 1} \right)\frac{\pi }{2} \cr & {\text{Now}},{\text{put }}\,n = - 2, - 1,0,1,21 \cr & \therefore x = \frac{{ - 3\pi }}{4},\frac{{ - \pi }}{4},\frac{\pi }{4},\frac{{5\pi }}{4}\,\,{\text{and}}\,\,\frac{{ - 3\pi }}{2},\frac{{ - \pi }}{2},\frac{\pi }{2},\frac{{3\pi }}{2},\frac{{5\pi }}{2} \cr & {\text{Since }}\, - \pi \leqslant x \leqslant \pi ,{\text{therefore}}, \cr & x = \pm \frac{\pi }{4}, \pm \frac{\pi }{2}, \pm \frac{{3\pi }}{4}{\text{only}}. \cr} $$

We have,
$$\eqalign{ & {\sin ^{50}}x - {\cos ^{50}}x = 1 \cr & \Rightarrow {\sin ^{50}}x = 1 + {\cos ^{50}}x \cr} $$
Since, $${\sin ^{50}}x \leqslant 1$$   and $$1 + {\cos ^{50}}x \geqslant 1$$    therefore, the two sides are equal only if
$$\eqalign{ & {\sin ^{50}}x = 1 = 1 + {\cos ^{50}}x{\text{ i}}{\text{.e }}{\sin ^{50}}x = 1\,{\text{and }}{\cos ^{50}}x = 0 \cr & \therefore x = 2n\pi + \frac{\pi }{2},n \in I. \cr} $$

$$\eqalign{ & {\text{They are in G}}{\text{.P}}{\text{., then}} \cr & \sin \,\alpha .\cos \,2\alpha = 1 \cr & \sin \,\alpha \left( {1 - 2\,{{\sin }^2}\alpha } \right) = 1 \cr & \sin \,\alpha - 2\,{\sin ^3}\alpha - 1 = 0 \cr & \sin \,\alpha = - 1{\text{ or }}\frac{{1 - i}}{2}{\text{ or }}\frac{{1 + i}}{2} \cr & {\text{So,}} \cr & \alpha = n\pi + {\left( { - 1} \right)^n}\frac{{ - \pi }}{2} \cr & \alpha = n\pi + {\left( { - 1} \right)^{n - 1}}\frac{\pi }{2} \cr} $$

$$\eqalign{ & \cos x + \cos 2x + \cos 3x + \cos 4x = 0 \cr & \Rightarrow \,\,2\cos 2x\cos x + 2\cos 3x\cos x = 0 \cr & \Rightarrow \,\,2\cos x\left( {2\cos \frac{{5x}}{2}\cos \frac{x}{2}} \right) = 0 \cr & \cos x = 0,\,\cos \frac{{5x}}{2} = 0,\,\,\cos \frac{x}{2} = 0 \cr & x = \pi ,\frac{\pi }{2},\frac{{3\pi }}{2},\frac{\pi }{5},\frac{{3\pi }}{5},\frac{{7\pi }}{5},\frac{{9\pi }}{5} \cr} $$

$$\eqalign{ & {\text{Given, }}\cos \theta + \cos 2\theta + \cos 3\theta = 0 \cr & \Rightarrow \left( {\cos 3\theta + \cos \theta } \right) + \cos 2\theta = 0 \cr & \Rightarrow 2\cos 2\theta \cdot \cos \theta + \cos 2\theta = 0 \cr & \Rightarrow \cos 2\theta \cdot \left( {2\cos \theta + 1} \right) = 0 \cr & {\text{we have}},\cos \theta = \cos \alpha \cr & \Rightarrow \theta = 2n\pi \pm \alpha \cr & \therefore {\text{For general value of }}\,\theta ,\cos 2\theta = 0 \cr & \Rightarrow \cos 2\theta = \cos \frac{\pi }{2} \cr & \Rightarrow 2\theta = 2m\pi \pm \frac{\pi }{2} \cr & \Rightarrow \theta = m\pi \pm \frac{\pi }{4}{\text{ or }}\,2\cos \theta + 1 = 0; \cr & \Rightarrow \cos \theta = \frac{{ - 1}}{2} \cr & \Rightarrow \cos \theta = \cos \frac{{2\pi }}{3} \cr & {\text{So}},\theta = 2m\pi \pm \frac{{2\pi }}{3} \cr} $$

$$\max \cos \theta = 1.$$    So, the equation can have solution only when $$\cos x = 1,\cos y = 1$$
$$\eqalign{ & \Rightarrow \,\,x = 0,y = 0 \cr & \Rightarrow \,\,\cos \left( {x - y} \right) = \cos {0^ \circ } = 1. \cr} $$