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Question: The general solution \(\cos \theta = \cos x\;\;{\text{is}}\;\;\theta = 2n\pi \pm x,n \in I\)? A.T...

The general solution cosθ=cosx    is    θ=2nπ±x,nI\cos \theta = \cos x\;\;{\text{is}}\;\;\theta = 2n\pi \pm x,n \in I?
A.True
B.False

Explanation

Solution

Hint : We should know about trigonometry properties to solve such type of problems. Some formulas of trigonometry are needed which are given as-
cos(A)cos(B)=2sin(A+B2)sin(AB2) sinθ=0θ=nπ  \cos \left( A \right) - \cos \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) \\\ \sin \theta = 0 \Rightarrow \theta = n\pi \\\

Complete step-by-step answer :
The given function iscosθ=cosx\cos \theta = \cos x.
The general solution for cosθ=cosx\cos \theta = \cos xis
cosθcosx=0\cos \theta - \cos x = 0
Applying formulacos(A)cos(B)=2sin(A+B2)sin(AB2)\cos \left( A \right) - \cos \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) then we get

2sin(θ+x2)sin(θx2)=0 sin(θ+x2)=0  or  sin(θx2)=0  2\sin \left( {\dfrac{{\theta + x}}{2}} \right)\sin \left( {\dfrac{{\theta - x}}{2}} \right) = 0 \\\ \sin \left( {\dfrac{{\theta + x}}{2}} \right) = 0\;{\text{or}}\;\sin \left( {\dfrac{{\theta - x}}{2}} \right) = 0 \\\

Further solving we get
(θ+x2)=nπ  or  (θx2)=nπ\left( {\dfrac{{\theta + x}}{2}} \right) = n\pi \;{\text{or}}\;\left( {\dfrac{{\theta - x}}{2}} \right) = n\pi
On taking the common solution from both the conditions
θ=2nπ±x,  where  nI\theta = 2n\pi \pm x,\;{\text{where}}\;n \in I
Thus the answer is true.
So, the correct answer is “Option A”.

Note : We should be aware of trigonometry formulas. The trigonometric functions are also called circular functions or angle functions. These are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. These functions are used in geometry, navigation, celestial mechanics, solid mechanics and many more. The most widely used trigonometric functions are the sine{\text{sine}}, the cosine{\text{cosine}}and the tangent.{\text{tangent}}{\text{.}}