Question

The general solution $\cos \theta = \cos x\;\;{\text{is}}\;\;\theta = 2n\pi \pm x,n \in I$?
A.True
B.False

Answer 1

Hint: We should know about trigonometry properties to solve such type of problems. Some formulas of trigonometry are needed which are given as-
$
  \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 is\[\cos \theta = \cos x\].
The general solution for \[\cos \theta = \cos x\]is
\[\cos \theta - \cos x = 0\]
Applying formula$\cos \left( A \right) - \cos \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ then we get
\[
  2\sin \left( {\dfrac{{\theta + x}}{2}} \right)\sin \left( {\dfrac{{\theta - x}}{2}} \right) = 0 \\
  \sin \left( {\dfrac{{\theta + x}}{2}} \right) = 0\;{\text{or}}\;\sin \left( {\dfrac{{\theta - x}}{2}} \right) = 0 \\
\]
Further solving we get
\[\left( {\dfrac{{\theta + x}}{2}} \right) = n\pi \;{\text{or}}\;\left( {\dfrac{{\theta - x}}{2}} \right) = n\pi \]
On taking the common solution from both the conditions
\[\theta = 2n\pi \pm x,\;{\text{where}}\;n \in I\]
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 ${\text{sine}}$, the ${\text{cosine}}$and the ${\text{tangent}}{\text{.}}$