
One root of the equation \[\cos x - x + \frac{1}{2} = 0\] lies in the interval
A. \[\left[ {0,\frac{\pi }{2}} \right]\]
В. \[\left[ { - \frac{\pi }{2},0} \right]\]
C. \[\left[ {\frac{\pi }{2},\pi } \right]\]
D. \[\left[ {\pi ,\frac{{3\pi }}{2}} \right]\]
Answer
164.4k+ views
HINT
When algebraic operations are used to solve a trigonometric problem, the term "root loss" describes how some roots are lost. Usually, it happens as both sides cancel out any identical phrases. The roots that are obtained when the terms in trigonometric equations are squared are known as extraneous roots. To accurately find the answer, it is necessary to check for extraneous roots.
Complete Step-by Step Solution:
We have been provided in the question about cos function that,
Let, \[f(x) = \cos x - x + \frac{1}{2} = 0\]
And we are to find the interval where the root lies in.
Now, we have to assume that suppose,
\[{\rm{x}} = 0\]
Then on substituting\[{\rm{x}} = 0\]on the equation above, we obtain
\[f(0) = \cos 0 - 0 + \frac{1}{2}\]
Now, we have to simplify the above equation, we obtain
\[ = \frac{1}{2} + \cos \left( 0 \right)\]
We have been already known that the x-coordinate (1) is equivalent to the value of
\[cos{0^\circ }\]
Therefore, we can write it as below
\[cos{0^\circ } = 1\]
On using the above trivial identity, the equation becomes
\[ = 1 + \frac{1}{2}\]
Now, we have to simplify the above expression to less complicated form:
Thus, we obtain
\[ = \frac{3}{2} > 0\]
Now replace the equation\[f(x) = \cos x - x + \frac{1}{2} = 0\]with\[{\rm{x}} = \frac{\pi }{2}\]
On substituting the value, the equation becomes
\[f\left( {\frac{\pi }{2}} \right) = \cos \frac{\pi }{2} - \frac{\pi }{2} + \frac{1}{2} = \frac{1}{2} - \frac{\pi }{2} < 0\]
Hence, the resultant will be obtained from the previous calculation as below,
\[f(0) > 0\]and\[f\left( {\frac{\pi }{2}} \right) < 0\]
This shows that \[{\rm{f}}\] changes its sign from\[\left( {0,\frac{\pi }{2}} \right)\].
This indicates that it contains at least one root\[\left( {0,\frac{\pi }{2}} \right)\] where\[f\]changes its sign from\[\left( {0,\frac{\pi }{2}} \right)\] in the interval\[\left( {0,\frac{\pi }{2}} \right)\].
Therefore, one root of the equation \[\cos x - x + \frac{1}{2} = 0\] lies in the interval
\[\left( {0,\frac{\pi }{2}} \right)\]
Hence, option A is correct
Note
When learning trigonometry, students frequently run across mistakes, misunderstandings, and barriers. While thinking about the cases, one should carefully analyze the regions. Students found it difficult while finding intervals. In general, just solve it without any limitations. Apply the constraint after that to your collection of solutions. The interval constraint may occasionally result in a novel solution that could be useful. Students are likely to make mistakes in these types of problems as it contains intervals which are sometimes tougher to solve.
When algebraic operations are used to solve a trigonometric problem, the term "root loss" describes how some roots are lost. Usually, it happens as both sides cancel out any identical phrases. The roots that are obtained when the terms in trigonometric equations are squared are known as extraneous roots. To accurately find the answer, it is necessary to check for extraneous roots.
Complete Step-by Step Solution:
We have been provided in the question about cos function that,
Let, \[f(x) = \cos x - x + \frac{1}{2} = 0\]
And we are to find the interval where the root lies in.
Now, we have to assume that suppose,
\[{\rm{x}} = 0\]
Then on substituting\[{\rm{x}} = 0\]on the equation above, we obtain
\[f(0) = \cos 0 - 0 + \frac{1}{2}\]
Now, we have to simplify the above equation, we obtain
\[ = \frac{1}{2} + \cos \left( 0 \right)\]
We have been already known that the x-coordinate (1) is equivalent to the value of
\[cos{0^\circ }\]
Therefore, we can write it as below
\[cos{0^\circ } = 1\]
On using the above trivial identity, the equation becomes
\[ = 1 + \frac{1}{2}\]
Now, we have to simplify the above expression to less complicated form:
Thus, we obtain
\[ = \frac{3}{2} > 0\]
Now replace the equation\[f(x) = \cos x - x + \frac{1}{2} = 0\]with\[{\rm{x}} = \frac{\pi }{2}\]
On substituting the value, the equation becomes
\[f\left( {\frac{\pi }{2}} \right) = \cos \frac{\pi }{2} - \frac{\pi }{2} + \frac{1}{2} = \frac{1}{2} - \frac{\pi }{2} < 0\]
Hence, the resultant will be obtained from the previous calculation as below,
\[f(0) > 0\]and\[f\left( {\frac{\pi }{2}} \right) < 0\]
This shows that \[{\rm{f}}\] changes its sign from\[\left( {0,\frac{\pi }{2}} \right)\].
This indicates that it contains at least one root\[\left( {0,\frac{\pi }{2}} \right)\] where\[f\]changes its sign from\[\left( {0,\frac{\pi }{2}} \right)\] in the interval\[\left( {0,\frac{\pi }{2}} \right)\].
Therefore, one root of the equation \[\cos x - x + \frac{1}{2} = 0\] lies in the interval
\[\left( {0,\frac{\pi }{2}} \right)\]
Hence, option A is correct
Note
When learning trigonometry, students frequently run across mistakes, misunderstandings, and barriers. While thinking about the cases, one should carefully analyze the regions. Students found it difficult while finding intervals. In general, just solve it without any limitations. Apply the constraint after that to your collection of solutions. The interval constraint may occasionally result in a novel solution that could be useful. Students are likely to make mistakes in these types of problems as it contains intervals which are sometimes tougher to solve.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
