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
Verified86.7k+ 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.
Last updated date: 21st Sep 2023
•
Total views: 86.7k
•
Views today: 0.86k
