Question

How many true solutions does the equation sine x equals cosine x minus 1 have over the interval 0 less than or equals x less than or equals 2 pi?
(1 point)
Responses

one
one

two
two

three
three

four

Answer 1

To solve the equation \(\sin x = \cos x - 1\), we can rearrange it:

\[
\sin x + 1 = \cos x
\]

Next, we can use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). We substitute \(\cos x\) from our rearranged equation:

\[
\sin^2 x + (\sin x + 1)^2 = 1
\]

Expanding the equation:

\[
\sin^2 x + (\sin^2 x + 2\sin x + 1) = 1
\]

Combining like terms gives:

\[
2\sin^2 x + 2\sin x + 1 = 1
\]

Subtracting 1 from both sides:

\[
2\sin^2 x + 2\sin x = 0
\]

Factoring out \(2\sin x\):

\[
2\sin x(\sin x + 1) = 0
\]

This implies two cases:

1. \(2\sin x = 0\) \(\Rightarrow \sin x = 0\)
2. \(\sin x + 1 = 0\) \(\Rightarrow \sin x = -1\)

Now, we find solutions for both cases in the interval \(0 \leq x \leq 2\pi\):

1. For \(\sin x = 0\):
- The solutions are \(x = 0, \pi, 2\pi\).

2. For \(\sin x = -1\):
- The solution is \(x = \frac{3\pi}{2}\).

Now, we list the solutions from both cases:

- From \(\sin x = 0\): \(x = 0, \pi, 2\pi\) (3 solutions)
- From \(\sin x = -1\): \(x = \frac{3\pi}{2}\) (1 solution)

In total, we have 4 solutions: \(0, \pi, 2\pi, \frac{3\pi}{2}\).

Thus, the number of true solutions to the equation \(\sin x = \cos x - 1\) in the interval \(0 \leq x \leq 2\pi\) is **four**.