Questions: Find all solutions of the equation in the interval [0,2π sqrt(1-sin x)=cos x

Solution Steps

Given the equation:

\[ \sqrt{1 - \sin x} = \cos x \]

Square both sides to eliminate the square root:

\[ 1 - \sin x = \cos^2 x \]

Use the identity \(\cos^2 x = 1 - \sin^2 x\) to rewrite the equation:

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

Rearrange the equation:

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

Factor the equation:

\[ \sin x (\sin x - 1) = 0 \]

Set each factor to zero:

1. \(\sin x = 0\)
2. \(\sin x = 1\)

Solve for \(x\) in the interval \([0, 2\pi]\):

1. \(\sin x = 0\) gives \(x = 0, \pi, 2\pi\)
2. \(\sin x = 1\) gives \(x = \frac{\pi}{2}\)

Check each solution in the original equation:

• \(x = 0\): \(\sqrt{1 - \sin 0} = \cos 0 \Rightarrow 1 = 1\) (Valid)
• \(x = \pi\): \(\sqrt{1 - \sin \pi} = \cos \pi \Rightarrow 1 \neq -1\) (Invalid)
• \(x = 2\pi\): \(\sqrt{1 - \sin 2\pi} = \cos 2\pi \Rightarrow 1 = 1\) (Valid)
• \(x = \frac{\pi}{2}\): \(\sqrt{1 - \sin \frac{\pi}{2}} = \cos \frac{\pi}{2} \Rightarrow 0 \neq 0\) (Invalid)

Final Answer