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Chapter 29: Problem 927

Find the solution set on \([0,2 \pi]\) for the equation \(\sin x \cos x=\cos x\)

Short Answer

Expert verified
The solution set for the given equation \(\sin x \cos x = \cos x\) within the interval [0, 2蟺] is \(x = \{ \frac{\pi}{2}, \frac{3\pi}{2} \}\).

Step by step solution

01

Simplify the equation

Firstly, we will factor out the common term, which in this case, is cos(x). We have: \(\sin x \cos x = \cos x\) Factoring out cos(x) from both sides, we get: \(\cos x[\sin x - 1] = 0\) Now, we need to find the x values in the interval [0, 2蟺] that would make either cos(x) = 0 or sin(x) - 1=0.
02

Solve cos(x) = 0

To solve for x values that would make cos(x) = 0, we need to determine where the cosine function equals zero within the specified interval. Recall that cos(x) = 0 at the angle 蟺/2 and 3蟺/2 within the interval [0, 2蟺]. Therefore, we have two possible solutions for this case: \(x_1 = \frac{\pi}{2}\) \(x_2 = \frac{3\pi}{2}\)
03

Solve sin(x) - 1 = 0

Now, we need to solve for the x values that would make sin(x) - 1 = 0. Adding 1 to both sides of the equation, we get: \(\sin x = 1\) Recall that sin(x) = 1 at the angle x = 蟺/2 within the interval [0, 2蟺]. However, this solution has already been found in the previous step. Thus, there is no new solution derived from this case.
04

Combine the solutions

As a final step, we need to combine the solutions obtained from Steps 2 and 3. The solution set for the given equation \(\sin x \cos x = \cos x\) within the interval [0, 2蟺] is: \(x = \{ \frac{\pi}{2}, \frac{3\pi}{2} \}\)

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