
If ${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x$ , then the number of solutions in the interval $[-2\pi ,2\pi ]$ is
(a) 1
(b) 2
(c) 3
(d) 4
Answer
626.4k+ views
Hint: Start by writing 1 as ${{\log }_{0.5}}0.5$ , followed by using the formula $\log a-\log b=\log \dfrac{a}{b}$ . Don’t forget to cross-check your result with the domain of log function.
Complete step-by-step answer:
Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.
Next, let us see the graph of cosx.
Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . We will now solve the equation given in the question.
${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x$
Now, we know 1 can be written as ${{\log }_{0.5}}0.5$ . So, using this in our equation, we get
${{\log }_{0.5}}\sin x={{\log }_{0.5}}0.5-{{\log }_{0.5}}\cos x$
We know $\log a-\log b=\log \dfrac{a}{b}$ , applying this to our equation, we get
${{\log }_{0.5}}\sin x={{\log }_{0.5}}\dfrac{0.5}{\cos x}$
Now we will remove the log from both sides provided that the part inside the log must be positive as the log function is defined only for positive real numbers. Therefore, sinx and cosx must be positive, providing the constraint that x must lie in the first quadrant as sinx and cosx are both positive only in the first quadrant.
$\sin x=\dfrac{0.5}{\cos x}$
$\Rightarrow 2\sin x\cos x=1$
Now using the formula sin2A=2sinAcosA, we get
$\sin 2x=1$
Sin2x is equal to 1 for $x={{\dfrac{\pi }{4}}},-\dfrac{7{{\pi}}}{4}$ in the given range of $[-2\pi ,2\pi ]$ and lie in the first and second quadrant.
Therefore, there are two solutions of the given equation in the range of $x\in [-2\pi ,2\pi ]$.
Hence, the answer is option (b).
Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two. Also, remember the property of the log functions along with its range and domain.
Complete step-by-step answer:
Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.
Next, let us see the graph of cosx.
Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . We will now solve the equation given in the question.
${{\log }_{0.5}}\sin x=1-{{\log }_{0.5}}\cos x$
Now, we know 1 can be written as ${{\log }_{0.5}}0.5$ . So, using this in our equation, we get
${{\log }_{0.5}}\sin x={{\log }_{0.5}}0.5-{{\log }_{0.5}}\cos x$
We know $\log a-\log b=\log \dfrac{a}{b}$ , applying this to our equation, we get
${{\log }_{0.5}}\sin x={{\log }_{0.5}}\dfrac{0.5}{\cos x}$
Now we will remove the log from both sides provided that the part inside the log must be positive as the log function is defined only for positive real numbers. Therefore, sinx and cosx must be positive, providing the constraint that x must lie in the first quadrant as sinx and cosx are both positive only in the first quadrant.
$\sin x=\dfrac{0.5}{\cos x}$
$\Rightarrow 2\sin x\cos x=1$
Now using the formula sin2A=2sinAcosA, we get
$\sin 2x=1$
Sin2x is equal to 1 for $x={{\dfrac{\pi }{4}}},-\dfrac{7{{\pi}}}{4}$ in the given range of $[-2\pi ,2\pi ]$ and lie in the first and second quadrant.
Therefore, there are two solutions of the given equation in the range of $x\in [-2\pi ,2\pi ]$.
Hence, the answer is option (b).
Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two. Also, remember the property of the log functions along with its range and domain.
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