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:
- \(2\sin x = 0\) \(\Rightarrow \sin x = 0\)
- \(\sin x + 1 = 0\) \(\Rightarrow \sin x = -1\)
Now, we find solutions for both cases in the interval \(0 \leq x \leq 2\pi\):
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For \(\sin x = 0\):
- The solutions are \(x = 0, \pi, 2\pi\).
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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.