
If $ {\log _{0.5}}(\sin x) = 1 - {\log _{0.5}}(\cos x) $ ,then the number of values of $ x \in [ - 2\pi ,2\pi ] $ is
Answer
560.7k+ views
Hint: The logarithmic function has some nice properties which we can use to simplify the given equation and reach at a point to obtain a simple trigonometric function where the general value of $ x $ can be found using simple trigonometric identity. Then we can identify which values of $ x \in [ - 2\pi ,2\pi ] $ .
$ {\log _a}(a) = 1 $
$ {\log _a}(b) = {\log _a}(c) \Rightarrow b = c $
$ {\log _a}(\dfrac{b}{c}) = {\log _a}(b) - {\log _a}(c) $ where, $ a,b,c \in \mathbb{R} $
$ 2\sin x\cos x = \sin 2x $
Complete step-by-step answer:
The given trigonometric equation is:
$ {\log _{0.5}}(\sin x) = 1 - {\log _{0.5}}(\cos x) $ --(1)
We need to solve the trigonometric equation.
Since, $ {\log _{0.5}}(0.5) = 1 $ so (1) can be modified as:
$ {\log _{0.5}}(\sin x) = {\log _{0.5}}(0.5) - {\log _{0.5}}(\cos x) $
$ \Rightarrow {\log _{0.5}}(\sin x) = {\log _{0.5}}(\dfrac{{0.5}}{{\cos x}}) $
$ \Rightarrow \sin x = \dfrac{{0.5}}{{\cos x}} $ [ $ \because {\log _a}(b) = {\log _a}(c) \Rightarrow b = c $ ]
On cross multiplication we get:
$ \Rightarrow \sin x\cos x = 0.5 $
$ \Rightarrow 2\sin x\cos x = 2 \times 0.5 $ [ Multiplying $ 2 $ on both sides of the equation]
$ \Rightarrow \sin 2x = 1 $
The general solution for the above equation is:
$ 2x = \dfrac{{(4n + 1)\pi }}{2} $ where, $ n \in \mathbb{Z} $
$ \Rightarrow x = \dfrac{{(4n + 1)\pi }}{4} $ where, $ n \in \mathbb{Z} $
For $ n = 0 $ ,
$ x = \dfrac{{(0 + 1)\pi }}{4} = \dfrac{\pi }{4} \in [ - 2\pi ,2\pi ] $
For $ n = 1 $ ,
$ x = \dfrac{{(4 + 1)\pi }}{4} = \dfrac{{5\pi }}{4} \in [ - 2\pi ,2\pi ] $
For $ n = 2 $ ,
$ x = \dfrac{{(8 + 1)\pi }}{4} = \dfrac{{9\pi }}{4} \notin [ - 2\pi ,2\pi ] $
Similarly, for $ n = - 1 $ ,
$ x = \dfrac{{( - 4 + 1)\pi }}{4} = \dfrac{{ - 3\pi }}{4} \in [ - 2\pi ,2\pi ] $
Similarly, for $ n = - 2 $ ,
$ x = \dfrac{{( - 8 + 1)\pi }}{4} = \dfrac{{ - 7\pi }}{4} \in [ - 2\pi ,2\pi ] $
Similarly, for $ n = - 3 $ ,
$ x = \dfrac{{( - 12 + 1)\pi }}{4} = \dfrac{{ - 11\pi }}{4} \notin [ - 2\pi ,2\pi ] $
Thus, the number of values of $ x \in [ - 2\pi ,2\pi ] $ is $ 4 $ which can be listed as $ \{ \dfrac{{ - 7\pi }}{4},\dfrac{{ - 3\pi }}{4},\dfrac{\pi }{4},\dfrac{{5\pi }}{4}\} $
Therefore, the number of values of $ x \in [ - 2\pi ,2\pi ] $ is $ 4 $ .
Note: Keep in mind that the general solution should be given first priority instead of the principal solution of the trigonometric function. The principal solution is the smallest solution satisfying the trigonometric function.
$ {\log _a}(a) = 1 $
$ {\log _a}(b) = {\log _a}(c) \Rightarrow b = c $
$ {\log _a}(\dfrac{b}{c}) = {\log _a}(b) - {\log _a}(c) $ where, $ a,b,c \in \mathbb{R} $
$ 2\sin x\cos x = \sin 2x $
Complete step-by-step answer:
The given trigonometric equation is:
$ {\log _{0.5}}(\sin x) = 1 - {\log _{0.5}}(\cos x) $ --(1)
We need to solve the trigonometric equation.
Since, $ {\log _{0.5}}(0.5) = 1 $ so (1) can be modified as:
$ {\log _{0.5}}(\sin x) = {\log _{0.5}}(0.5) - {\log _{0.5}}(\cos x) $
$ \Rightarrow {\log _{0.5}}(\sin x) = {\log _{0.5}}(\dfrac{{0.5}}{{\cos x}}) $
$ \Rightarrow \sin x = \dfrac{{0.5}}{{\cos x}} $ [ $ \because {\log _a}(b) = {\log _a}(c) \Rightarrow b = c $ ]
On cross multiplication we get:
$ \Rightarrow \sin x\cos x = 0.5 $
$ \Rightarrow 2\sin x\cos x = 2 \times 0.5 $ [ Multiplying $ 2 $ on both sides of the equation]
$ \Rightarrow \sin 2x = 1 $
The general solution for the above equation is:
$ 2x = \dfrac{{(4n + 1)\pi }}{2} $ where, $ n \in \mathbb{Z} $
$ \Rightarrow x = \dfrac{{(4n + 1)\pi }}{4} $ where, $ n \in \mathbb{Z} $
For $ n = 0 $ ,
$ x = \dfrac{{(0 + 1)\pi }}{4} = \dfrac{\pi }{4} \in [ - 2\pi ,2\pi ] $
For $ n = 1 $ ,
$ x = \dfrac{{(4 + 1)\pi }}{4} = \dfrac{{5\pi }}{4} \in [ - 2\pi ,2\pi ] $
For $ n = 2 $ ,
$ x = \dfrac{{(8 + 1)\pi }}{4} = \dfrac{{9\pi }}{4} \notin [ - 2\pi ,2\pi ] $
Similarly, for $ n = - 1 $ ,
$ x = \dfrac{{( - 4 + 1)\pi }}{4} = \dfrac{{ - 3\pi }}{4} \in [ - 2\pi ,2\pi ] $
Similarly, for $ n = - 2 $ ,
$ x = \dfrac{{( - 8 + 1)\pi }}{4} = \dfrac{{ - 7\pi }}{4} \in [ - 2\pi ,2\pi ] $
Similarly, for $ n = - 3 $ ,
$ x = \dfrac{{( - 12 + 1)\pi }}{4} = \dfrac{{ - 11\pi }}{4} \notin [ - 2\pi ,2\pi ] $
Thus, the number of values of $ x \in [ - 2\pi ,2\pi ] $ is $ 4 $ which can be listed as $ \{ \dfrac{{ - 7\pi }}{4},\dfrac{{ - 3\pi }}{4},\dfrac{\pi }{4},\dfrac{{5\pi }}{4}\} $
Therefore, the number of values of $ x \in [ - 2\pi ,2\pi ] $ is $ 4 $ .
Note: Keep in mind that the general solution should be given first priority instead of the principal solution of the trigonometric function. The principal solution is the smallest solution satisfying the trigonometric function.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

