Find number of value of $x\in \left[ -2\pi ,2\pi \right]$, If ${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x$.
Last updated date: 20th Mar 2023
•
Total views: 305.7k
•
Views today: 4.84k
Answer
305.7k+ views
Hint: Use property of logarithm that is \[{{\log }_{a}}a=1\]and ${{\log }_{c}}a-{{\log }_{c}}b={{\log }_{c}}\left( \dfrac{a}{b} \right)$to simplify the given relation. Use a graphical approach to find values of ‘x’ for simplicity.
Complete step-by-step answer:
Here, it is given that ${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x$, and then we need to determine all the values of x lying in$\left[ -2\pi ,2\pi \right]$.
We have
${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x\ldots \ldots (1)$
As we know the property of the logarithm function that \[{{\log }_{a}}a=1\] where $a>0$and $a\ne 1$.
Or vice-versa is also true. It means we can replace ‘1’ from equation (1) by \[{{\log }_{0.5}}0.5\]for the simplification of the problem.
Hence, equation (1) can be written as
${{\log }_{0.5}}\sin x={{\log }_{0.5}}0.5-{{\log }_{0.5}}\cos x$
We can use property of logarithm ${{\log }_{c}}a-{{\log }_{c}}b={{\log }_{c}}\left( \dfrac{a}{b} \right)$, with the above equation and get
${{\log }_{0.5}}\sin x={{\log }_{0.5}}\left( \dfrac{0.5}{\cos x} \right)\ldots \ldots (2)$
As we know that ‘a’ should be equal to ‘b’ if ${{\log }_{c}}a={{\log }_{c}}b$.
Hence, using the above property with equation (2), we get
$\dfrac{\sin x}{1}=\dfrac{0.5}{\cos x}$
On cross-multiplying, we get
$\sin x\cos x=\dfrac{1}{2}$or $2\sin x\cos x=1\ldots \ldots (3)$
As we know the trigonometric identity of $\sin 2x$as $\sin 2x=2\sin x\cos x$or vice-versa.
Hence, equation (3) can be given as
$\sin 2x=1\ldots \ldots (4)$
Now, we have to find ‘x’ in the interval$\left[ -2\pi ,2\pi \right]$.
So we have $-2\pi \le x\le 2\pi $
Multiplying by ‘2’ on each side we get
$-4\pi \le 2x\le 4\pi $
Now, drawing graph of $\sin x$ from $-4\pi $ to \[4\pi \], we get
As we can observe that $y=\sin x$ has values of 1 at $\dfrac{\pi }{2},\dfrac{5\pi }{2},\dfrac{-3\pi }{2},\dfrac{-7\pi }{2}$.
Now, we have the equation $\sin 2x=1$.
Hence,
$2x=\dfrac{-7\pi }{2},\dfrac{-3\pi }{2},\dfrac{\pi }{2}.\dfrac{5\pi }{2}$
Or $x=\dfrac{-7\pi }{4},\dfrac{-3\pi }{4},\dfrac{\pi }{4}.\dfrac{5\pi }{4}$
Note: One can get confusion between $y=\sin x$ and equation $\sin 2x=1$. Graph of $y=\sin x$ is representing the general relation between angles and values which is not related to equation$\sin 2x=1$. One can suppose ‘2x’ as ‘t’ as well for the simplicity, so we will get equation $\sin t=1$. Now, t will lie in $\left[ -4\pi ,4\pi \right]$ as \[t=2x\]; hence find all values of ‘t’ then find ‘x’ by using relation $x=\dfrac{t}{2}$.
Complete step-by-step answer:
Here, it is given that ${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x$, and then we need to determine all the values of x lying in$\left[ -2\pi ,2\pi \right]$.
We have
${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x\ldots \ldots (1)$
As we know the property of the logarithm function that \[{{\log }_{a}}a=1\] where $a>0$and $a\ne 1$.
Or vice-versa is also true. It means we can replace ‘1’ from equation (1) by \[{{\log }_{0.5}}0.5\]for the simplification of the problem.
Hence, equation (1) can be written as
${{\log }_{0.5}}\sin x={{\log }_{0.5}}0.5-{{\log }_{0.5}}\cos x$
We can use property of logarithm ${{\log }_{c}}a-{{\log }_{c}}b={{\log }_{c}}\left( \dfrac{a}{b} \right)$, with the above equation and get
${{\log }_{0.5}}\sin x={{\log }_{0.5}}\left( \dfrac{0.5}{\cos x} \right)\ldots \ldots (2)$
As we know that ‘a’ should be equal to ‘b’ if ${{\log }_{c}}a={{\log }_{c}}b$.
Hence, using the above property with equation (2), we get
$\dfrac{\sin x}{1}=\dfrac{0.5}{\cos x}$
On cross-multiplying, we get
$\sin x\cos x=\dfrac{1}{2}$or $2\sin x\cos x=1\ldots \ldots (3)$
As we know the trigonometric identity of $\sin 2x$as $\sin 2x=2\sin x\cos x$or vice-versa.
Hence, equation (3) can be given as
$\sin 2x=1\ldots \ldots (4)$
Now, we have to find ‘x’ in the interval$\left[ -2\pi ,2\pi \right]$.
So we have $-2\pi \le x\le 2\pi $
Multiplying by ‘2’ on each side we get
$-4\pi \le 2x\le 4\pi $
Now, drawing graph of $\sin x$ from $-4\pi $ to \[4\pi \], we get

As we can observe that $y=\sin x$ has values of 1 at $\dfrac{\pi }{2},\dfrac{5\pi }{2},\dfrac{-3\pi }{2},\dfrac{-7\pi }{2}$.
Now, we have the equation $\sin 2x=1$.
Hence,
$2x=\dfrac{-7\pi }{2},\dfrac{-3\pi }{2},\dfrac{\pi }{2}.\dfrac{5\pi }{2}$
Or $x=\dfrac{-7\pi }{4},\dfrac{-3\pi }{4},\dfrac{\pi }{4}.\dfrac{5\pi }{4}$
Note: One can get confusion between $y=\sin x$ and equation $\sin 2x=1$. Graph of $y=\sin x$ is representing the general relation between angles and values which is not related to equation$\sin 2x=1$. One can suppose ‘2x’ as ‘t’ as well for the simplicity, so we will get equation $\sin t=1$. Now, t will lie in $\left[ -4\pi ,4\pi \right]$ as \[t=2x\]; hence find all values of ‘t’ then find ‘x’ by using relation $x=\dfrac{t}{2}$.
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
