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Question: If \( {\log _{0.5}}(\sin x) = 1 - {\log _{0.5}}(\cos x) \) ,then the number of values of \( x \in [ ...

If log⁡0.5(sin⁡x)=1−log⁡0.5(cos⁡x){\log _{0.5}}(\sin x) = 1 - {\log _{0.5}}(\cos x) ,then the number of values of x∈[−2π,2π]x \in [ - 2\pi ,2\pi ] is

Explanation

Solution

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 xx can be found using simple trigonometric identity. Then we can identify which values of x∈[−2π,2π]x \in [ - 2\pi ,2\pi ] .
log⁡a(a)=1{\log _a}(a) = 1
log⁡a(b)=log⁡a(c)⇒b=c{\log _a}(b) = {\log _a}(c) \Rightarrow b = c
log⁡a(bc)=log⁡a(b)−log⁡a(c){\log _a}(\dfrac{b}{c}) = {\log _a}(b) - {\log _a}(c) where, a,b,c∈Ra,b,c \in \mathbb{R}
2sin⁡xcos⁡x=sin⁡2x2\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){\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{\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){\log _{0.5}}(\sin x) = {\log _{0.5}}(0.5) - {\log _{0.5}}(\cos x)
⇒log⁡0.5(sin⁡x)=log⁡0.5(0.5cos⁡x)\Rightarrow {\log _{0.5}}(\sin x) = {\log _{0.5}}(\dfrac{{0.5}}{{\cos x}})
⇒sin⁡x=0.5cos⁡x\Rightarrow \sin x = \dfrac{{0.5}}{{\cos x}} [ ∵log⁡a(b)=log⁡a(c)⇒b=c\because {\log _a}(b) = {\log _a}(c) \Rightarrow b = c ]
On cross multiplication we get:
⇒sin⁡xcos⁡x=0.5\Rightarrow \sin x\cos x = 0.5
⇒2sin⁡xcos⁡x=2×0.5\Rightarrow 2\sin x\cos x = 2 \times 0.5 [ Multiplying 22 on both sides of the equation]
⇒sin⁡2x=1\Rightarrow \sin 2x = 1
The general solution for the above equation is:
2x=(4n+1)π22x = \dfrac{{(4n + 1)\pi }}{2} where, n∈Zn \in \mathbb{Z}
⇒x=(4n+1)π4\Rightarrow x = \dfrac{{(4n + 1)\pi }}{4} where, n∈Zn \in \mathbb{Z}
For n=0n = 0 ,
x=(0+1)π4=π4∈[−2π,2π]x = \dfrac{{(0 + 1)\pi }}{4} = \dfrac{\pi }{4} \in [ - 2\pi ,2\pi ]
For n=1n = 1 ,
x=(4+1)π4=5π4∈[−2π,2π]x = \dfrac{{(4 + 1)\pi }}{4} = \dfrac{{5\pi }}{4} \in [ - 2\pi ,2\pi ]
For n=2n = 2 ,
x=(8+1)π4=9π4∉[−2π,2π]x = \dfrac{{(8 + 1)\pi }}{4} = \dfrac{{9\pi }}{4} \notin [ - 2\pi ,2\pi ]
Similarly, for n=−1n = - 1 ,
x=(−4+1)π4=−3π4∈[−2π,2π]x = \dfrac{{( - 4 + 1)\pi }}{4} = \dfrac{{ - 3\pi }}{4} \in [ - 2\pi ,2\pi ]
Similarly, for n=−2n = - 2 ,
x=(−8+1)π4=−7π4∈[−2π,2π]x = \dfrac{{( - 8 + 1)\pi }}{4} = \dfrac{{ - 7\pi }}{4} \in [ - 2\pi ,2\pi ]
Similarly, for n=−3n = - 3 ,
x=(−12+1)π4=−11π4∉[−2π,2π]x = \dfrac{{( - 12 + 1)\pi }}{4} = \dfrac{{ - 11\pi }}{4} \notin [ - 2\pi ,2\pi ]
Thus, the number of values of x∈[−2π,2π]x \in [ - 2\pi ,2\pi ] is 44 which can be listed as −7π4,−3π4,π4,5π4\\{ \dfrac{{ - 7\pi }}{4},\dfrac{{ - 3\pi }}{4},\dfrac{\pi }{4},\dfrac{{5\pi }}{4}\\}
Therefore, the number of values of x∈[−2π,2π]x \in [ - 2\pi ,2\pi ] is 44 .

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.