Question

If $$ - \pi \leqslant x \leqslant \pi , - \pi \leqslant y \leqslant \pi $$     and $$\cos x + \cos y = 2$$    then the value $$\cos \left( {x - y} \right)$$   is

A. $$- 1$$
B. $$0$$
C. $$1$$  
D. None of these
Answer :   $$1$$
Solution :
$$\max \cos \theta = 1.$$    So, the equation can have solution only when $$\cos x = 1,\cos y = 1$$
$$\eqalign{ & \Rightarrow \,\,x = 0,y = 0 \cr & \Rightarrow \,\,\cos \left( {x - y} \right) = \cos {0^ \circ } = 1. \cr} $$

Releted MCQ Question on
Trigonometry >> Trignometric Equations

Releted Question 1

The equation $$2\,{\cos ^2}\frac{x}{2}{\sin ^2}x = {x^2} + {x^{ - 2}};0 < x \leqslant \frac{\pi }{2}$$        has

A. no real solution
B. one real solution
C. more than one solution
D. none of these
Releted Question 2

The general solution of the trigonometric equation $$\sin x + \cos x = 1$$    is given by:

A. $$x = 2n\pi ;\,\,n = 0,\,\, \pm 1,\,\, \pm 2\,.....$$
B. $$x = 2n\pi + \frac{\pi }{2};\,\,n = 0,\,\, \pm 1,\,\, \pm 2\,.....$$
C. $$x = n\pi + {\left( { - 1} \right)^n}\,\,\frac{\pi }{4} - \frac{\pi }{4}$$
D. none of these
Releted Question 3

The general solution of $$\sin \,x - 3\,\sin \,2x\, + \sin \,3x\, = \cos x - 3\,\cos \,\,2x + \cos \,3x$$           is

A. $$n\pi + \frac{\pi }{8}$$
B. $$\frac{{n\pi }}{2} + \frac{\pi }{8}$$
C. $${\left( { - 1} \right)^n}\frac{{n\pi }}{2} + \frac{\pi }{8}$$
D. $$2n\pi + {\cos ^{ - 1}}\frac{3}{2}$$
Releted Question 4

Number of solutions of the equation $$\tan x + \sec x = 2\cos x$$     lying in the interval $$\left[ {0,2\pi } \right]$$  is:

A. 0
B. 1
C. 2
D. 3

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