Trignometric Equations MCQ Questions & Answers in Trigonometry | Maths

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

$$\eqalign{ & \sqrt 3 \,\sec \,x\, + \,{\text{cosec}}\,x\, + 2\,\left( {\tan \,x\, - \cot \,x} \right) = 0 \cr & \Rightarrow \,\,\frac{{\sqrt 3 }}{2}\,\sin \,x\, + \frac{1}{2}\cos \,x\,\, = {\cos ^2}x - \,{\sin ^2}x \cr & \Rightarrow \,\,\cos \,\left( {x\, - \frac{\pi }{3}} \right)\, = \cos \,2x \cr & \Rightarrow \,\,x - \frac{\pi }{3}\, = 2n\pi \pm 2x \cr & \Rightarrow \,\,x = \frac{{2n\pi }}{3} + \frac{\pi }{9}\,\,{\text{or}}\,\,x = \, - 2n\pi \, - \frac{\pi }{3} \cr & {\text{For}}\,x\, \in \,S{\text{,}}\,n = {\text{0}} \cr & \Rightarrow \,\,x = \frac{\pi }{9},\, - \frac{\pi }{3} \cr & n = 1 \cr & \Rightarrow \,\,x = \frac{{7\pi }}{9};\, \cr & n = - 1 \cr & \Rightarrow \,\,x = \frac{{ - 5\pi }}{9} \cr} $$
∴ Sum of all values of $$x = \frac{\pi }{9} - \frac{\pi }{3} + \frac{{7\pi }}{9} - \frac{{5\pi }}{9} = 0$$

$$\eqalign{ & \sin 8\theta + \sin 2\theta = \sin 16\theta + \sin 2\theta \,\,\,{\text{or, }}\sin 16\theta = \sin 8\theta \cr & \therefore \,\,16\theta = n\pi + {\left( { - 1} \right)^n}8\theta \cr & \Rightarrow \,\,8\theta = 2m\pi ,\,{\text{when }}n\,{\text{is even}} \cr & 24\theta = \left( {2m + 1} \right)\pi ,\,{\text{when }}n\,{\text{is odd}} \cr & \therefore \,\,\theta = \frac{{m\pi }}{4},\frac{{\left( {2m + 1} \right)\pi }}{{24}},\,{\text{when }}m \in {\Bbb Z} \cr & \theta = 0,\frac{\pi }{4},\frac{\pi }{2}\,{\text{and }}\frac{\pi }{{24}},\frac{\pi }{8},\frac{{5\pi }}{{24}},\frac{{7\pi }}{{24}},\frac{{3\pi }}{8},\frac{{11\pi }}{{24}}. \cr} $$

$$\eqalign{ & {\text{We have}},\,\,\sin \pi \left( {{x^2} + x} \right) = \sin \pi {x^2} \cr & \Rightarrow \pi \left( {{x^2} + x} \right) = n\pi + {\left( { - 1} \right)^n}\pi {x^2} \cr & \therefore {\text{Either }}{x^2} + x = 2m + {x^2} \cr & \Rightarrow x = 2m \in I \cr & {\text{or }}{x^2} + x = k - {x^2},\,{\text{where }}k\,{\text{is an odd integer}} \cr & \Rightarrow 2{x^2} + x - k = 0 \cr & \Rightarrow x = \frac{{ - 1 \pm \sqrt {1 + 8k} }}{4} \cr} $$
For least positive non-integral solution is $$x = \frac{1}{2},{\text{when }}\,k = 1$$

$$\eqalign{ & \sin x + \cos x = 1 \cr & \Rightarrow \,\,\frac{1}{{\sqrt 2 }}\sin x + \frac{1}{{\sqrt 2 }}\cos = \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \,\,\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4} = \sin \frac{\pi }{4} \cr & \Rightarrow \,\,\sin \left( {x + \frac{\pi }{4}} \right) = \sin \frac{\pi }{4} \cr & \Rightarrow \,\,x + \frac{\pi }{4} = n\pi + {\left( { - 1} \right)^n}\frac{\pi }{4},n \in Z\left( {{\text{the set of integers}}} \right) \cr & \Rightarrow \,\,x = n\pi + {\left( { - 1} \right)^n}\frac{\pi }{4} - \frac{\pi }{4} \cr & {\text{where }}n = 0, \pm 1, \pm 2,..... \cr} $$

We know that
$$ \Rightarrow \,\, - \sqrt {{a^2} + {b^2}} \leqslant a\cos \theta + b\sin \theta \leqslant \sqrt {{a^2} + {b^2}} $$           NOTE THIS STEP
$$\eqalign{ & \Rightarrow \,\, - \sqrt {74} \leqslant 7\cos x + 5\sin x \leqslant \sqrt {74} \cr & \Rightarrow \,\, - \sqrt {74} \leqslant 2k + 1 \leqslant \sqrt {74} \cr & \Rightarrow \,\, - 8.6 \leqslant 2k + 1 \leqslant 8.6 \cr & \Rightarrow \,\, - 4.8 \leqslant k \leqslant 3.8 \cr} $$
(considering only integral values)
⇒ $$k$$ can take 8 integral values.

$$\eqalign{ & \sin \,x + 2\sin \,2x - \sin \,3x = 3 \cr & \Rightarrow \,\,\sin \,x + 4\sin \,x\,\cos \,x - 3\sin \,x + 4\,{\sin ^3}\,x = 3 \cr & \Rightarrow \,\,\sin \,x\,\left( { - 2 + 2\,\cos \,x + 4\,{{\sin }^2}x} \right) = 3 \cr & \Rightarrow \,\,\sin \,x\,\left( { - 2 + 2\,\cos \,x\, + 4 - 4\,{{\cos }^2}x} \right) = 3 \cr & \Rightarrow \,\,2 + 2\,\cos \,x - 4\,{\cos ^2}x = \frac{3}{{\sin \,x}} \cr & \Rightarrow \,\,2 - \,\,\left( {4\,{{\cos }^2}\,x - 2.\,\,2\,\cos \,x.\frac{1}{2} + \frac{1}{4}} \right)\,\, + \,\frac{1}{4} = \frac{3}{{\sin \,x}} \cr & \Rightarrow \,\,\frac{9}{4} - {\left( {2\,\cos \,x\, - \frac{1}{2}} \right)^2} = \frac{3}{{\sin \,x}} \cr & {\text{LHS}}\, \leqslant \,\frac{9}{4}\,{\text{and}}\,{\text{RHS}}\, \geqslant 3 \cr & \therefore \,\,{\text{Equation}}\,{\text{has}}\,{\text{no}}\,{\text{solution}} \cr} $$

The given equation is
$$\eqalign{ & \sin \,x - 3\,\sin \,2x\, + \sin \,3x\, = \cos x - 3\,\cos \,\,2x + \cos \,3x \cr & \Rightarrow \,2\,\sin \,2x\,\cos \,x - 3\,\sin \,2x = 2\,\cos \,\,2x\,\cos \,x - 3\,\cos \,\,2x \cr & \Rightarrow \,\sin \,2x\,\left( {2\,\cos \,x - 3} \right) = \cos \,\,2x\,\left( {2\,\cos \,x - 3} \right) \cr & \Rightarrow \,\sin \,2x = \cos \,2x\,\,\,\,\left( {{\text{as}}\,\cos \,x\, \ne \,\frac{3}{2}} \right) \cr & \Rightarrow \,\tan \,2x\, = 1\, \cr & \Rightarrow \,2x = n\,\pi \,\, + \,\frac{\pi }{4} \cr & \Rightarrow \,x = \frac{{n\pi }}{2} + \frac{\pi }{8} \cr} $$

The given equation can be written as
$$\eqalign{ & {e^{\sin x}} = 4 + \frac{1}{{{e^{\sin x}}}}\,\,\,.....\left( 1 \right) \cr & {\text{Now, }} - 1 \leqslant \sin x \leqslant 1\,{\text{and }}e < 3 \cr & \Rightarrow {e^{\sin x}} < 3 \cr & \Rightarrow {\text{Again as we always have }}\frac{1}{{{e^{\sin x}}}} > 0 \cr & \therefore 4 + \frac{1}{{{e^{\sin x}}}} > 4 \cr} $$
Thus the L.H.S. of $$\left( 1 \right) < 3$$  and R.H.S. of $$\left( 1 \right) > 4.$$
Hence there is no real values of $$x$$ which satisfy $$\left( 1 \right).$$
It follows that the given equation has no real solution.

Here, $$2{\sin ^2}x + \sin x - 1 = 0$$
$$ \Rightarrow \,\,\sin x = \frac{1}{2}, - 1.$$
But $$\sin x = - 1$$
$$ \Rightarrow \,\,x = \frac{{3\pi }}{2}$$
which does not satisfy the equation.