Solving Logarithmic and Trigonometric Equations

The question asks for the value of \(x\) that satisfies the equation \(log_{cos x} ~sin x = 1\) under the condition \(0 < x < \dfrac{\pi}{2}\).

Understanding the Logarithmic Equation

The given equation is in the form \(log_b a = c\), where:

• Base \(b = cos x\)
• Argument \(a = sin x\)
• Value \(c = 1\)

By the definition of logarithms, the equation \(log_b a = c\) is equivalent to \(b^c = a\).

Applying the Definition to Solve for x

Applying the definition to our equation \(log_{cos x} ~sin x = 1\), we get:

\[(cos x)^1 = sin x\]

This simplifies to:

\[cos x = sin x\]

Solving the Trigonometric Equation

We need to find the value of \(x\) in the interval \(0 < x < \dfrac{\pi}{2}\) for which \(cos x = sin x\).

In the interval \(0 < x < \dfrac{\pi}{2}\), both \(sin x\) and \(cos x\) are positive. Also, \(cos x\) is non-zero in this interval. Therefore, we can divide both sides by \(cos x\):

\[\dfrac{sin x}{cos x} = 1\]

Recall that \(\dfrac{sin x}{cos x} = tan x\).

So, the equation becomes:

\[tan x = 1\]

Finding the Value of x in the Given Range

We need to find the value of \(x\) in the range \(0 < x < \dfrac{\pi}{2}\) such that \(tan x = 1\). The angle in this range whose tangent is 1 is \(\dfrac{\pi}{4}\).

So, \(x = \dfrac{\pi}{4}\).

Verifying the Solution and Logarithm Domain

We must check if \(x = \dfrac{\pi}{4}\) satisfies the original equation and the domain requirements for the logarithm \(log_{cos x} ~sin x = 1\). The domain requirements for \(log_b a\) are \(b > 0\), \(b \neq 1\), and \(a > 0\).

For \(x = \dfrac{\pi}{4}\):

• Base: \(cos x = cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}\)
• Argument: \(sin x = sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}\)

Let's check the domain requirements for the base \(cos x\):

• \(cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}\). Since \(\dfrac{1}{\sqrt{2}} \approx 0.707\), it is greater than 0. (Requirement \(b > 0\) is satisfied).
• \(cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}\). Since \(\dfrac{1}{\sqrt{2}} \neq 1\). (Requirement \(b \neq 1\) is satisfied).

Let's check the domain requirements for the argument \(sin x\):

• \(sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}\). Since \(\dfrac{1}{\sqrt{2}} > 0\). (Requirement \(a > 0\) is satisfied).

All domain requirements are satisfied for \(x = \dfrac{\pi}{4}\). Also, the range \(0 < \dfrac{\pi}{4} < \dfrac{\pi}{2}\) is satisfied.

Substituting \(x = \dfrac{\pi}{4}\) back into the original equation:

\[log_{cos (\pi/4)} ~sin (\pi/4) = log_{1/\sqrt{2}} (1/\sqrt{2})\]

Since \(log_b b = 1\), we have \(log_{1/\sqrt{2}} (1/\sqrt{2}) = 1\). The equation is satisfied.

Therefore, the value of \(x\) satisfying the given equation and condition is \(\dfrac{\pi}{4}\).

Revision Table: Key Concepts

Understanding the core concepts used in this problem is crucial for solving similar equations.

Additional Information: Solving Trigonometric Equations

When solving trigonometric equations like \(tan x = 1\), remember that tangent has a period of \(\pi\). The general solution for \(tan x = 1\) is \(x = n\pi + \dfrac{\pi}{4}\), where \(n\) is an integer.

However, the problem specifies the interval \(0 < x < \dfrac{\pi}{2}\). Within this specific interval, there is only one solution:

• For \(n=0\), \(x = 0 \cdot \pi + \dfrac{\pi}{4} = \dfrac{\pi}{4}\). This is within \(0 < x < \dfrac{\pi}{2}\).
• For \(n=1\), \(x = 1 \cdot \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}\). This is outside \(0 < x < \dfrac{\pi}{2}\).
• For \(n=-1\), \(x = -1 \cdot \pi + \dfrac{\pi}{4} = -\dfrac{3\pi}{4}\). This is outside \(0 < x < \dfrac{\pi}{2}\).

Thus, within the given constraint \(0 < x < \dfrac{\pi}{2}\), the only solution is \(x = \dfrac{\pi}{4}\).

Always remember to check the domain constraints imposed by logarithmic or other functions present in the original equation, as not all solutions from the simplified trigonometric equation might be valid for the original problem.