The smallest positive root of the trigonometric equation √(sin(1-x)=√(cosx) is equal to what?

To find the smallest positive root of the equation \( \sqrt{\sin(1 - x)} = \sqrt{\cos x} \), we can start by squaring both sides to eliminate the square roots. This gives us the equation \( \sin(1 - x) = \cos x \). From there, we can use some trigonometric identities and properties to help us solve for \( x \).

Setting Up the Equation

First, we rewrite the equation as follows:

\( \sin(1 - x) = \cos x \)

Using the identity \( \cos x = \sin\left(\frac{\pi}{2} - x\right) \), we can transform this equation into:

\( \sin(1 - x) = \sin\left(\frac{\pi}{2} - x\right) \)

Using the Sine Identity

For two sine functions to be equal, we have two main cases to consider:

• Case 1: \( 1 - x = \frac{\pi}{2} - x + 2k\pi \) for some integer \( k \)
• Case 2: \( 1 - x = \pi - \left(\frac{\pi}{2} - x\right) + 2k\pi \) for some integer \( k \)

Solving Case 1

In Case 1, we can simplify:

\( 1 = \frac{\pi}{2} + 2k\pi \)

Rearranging gives:

\( 2k\pi = 1 - \frac{\pi}{2} \)

\( 2k\pi = \frac{2 - \pi}{2} \)

This leads to:

\( k = \frac{2 - \pi}{4\pi} \)

Since \( k \) must be an integer, we can see that \( k = 0 \) is the only feasible option here, giving us:

\( 1 = \frac{\pi}{2} \), which is not valid.

Solving Case 2

Now, let’s explore Case 2:

\( 1 - x = \pi - \left(\frac{\pi}{2} - x\right) + 2k\pi \)

This simplifies to:

\( 1 - x = \pi - \frac{\pi}{2} + x + 2k\pi \)

\( 1 - x = \frac{\pi}{2} + x + 2k\pi \)

Rearranging gives:

\( 1 - \frac{\pi}{2} = 2x + 2k\pi \)

\( 2x = 1 - \frac{\pi}{2} - 2k\pi \)

Thus:

\( x = \frac{1 - \frac{\pi}{2} - 2k\pi}{2} \)

Finding the Smallest Positive Root

To find the smallest positive root, we can set \( k = 0 \):

\( x = \frac{1 - \frac{\pi}{2}}{2} \)

Calculating this gives us:

\( x = \frac{1}{2} - \frac{\pi}{4} \)

Evaluating the Result

Next, we need to evaluate whether this value is positive. Since \( \frac{\pi}{4} \approx 0.785 \) and \( \frac{1}{2} = 0.5 \), we find that:

\( \frac{1}{2} - \frac{\pi}{4} \) is negative. Therefore, we should try \( k = -1 \) instead:

\( x = \frac{1 - \frac{\pi}{2} + 2\pi}{2} = \frac{1 + \frac{3\pi}{2}}{2} \)

This will yield a positive value.

Ultimately, the smallest positive root of the equation \( \sqrt{\sin(1 - x)} = \sqrt{\cos x} \) is approximately \( 1.5 \). So, we conclude that the smallest positive solution is:

x ≈ 1.5