A student is asked to solve the equation sin2x=1\sin^2 x = 1sin2x=1 for 0°⩽x⩽360°0° \leqslant x \leqslant 360°0°⩽x⩽360°.
The student writes:
Step 1: sinx=1\sin x = 1sinx=1
Step 2: x=90°x = 90°x=90°
Explain the error the student has made in Step 1.
State the correct solutions of sin2x=1\sin^2 x = 1sin2x=1 for 0°⩽x⩽360°0° \leqslant x \leqslant 360°0°⩽x⩽360°.
A second student solves the equation sinxcosx=sinx\sin x\cos x = \sin xsinxcosx=sinx for 0°⩽x<360° 0° \leqslant x < 360°\,0°⩽x<360° by dividing both sides by sinx\sin xsinx, obtaining cosx=1\cos x = 1cosx=1 and hence x=0°x = 0°x=0°. Explain what is wrong with this method, and give the complete solution set.