Question

Find all solutions of sin^3(x) +s sin^2(x) - sinx -1 = 0 on the interval [0, 2π]

Answer 1

\[\sin^3x + \sin^2x -sinx -1 =0\]
=\[(\sin^2x-1)(sinx+1)\]
\[\sin^2x=1 | sinx = -1\]
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\[sinx = -1, x=3 \pi/2\]
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\[\sin^2x = 1 , sinx = \pm1\]
\[x=\pi/2\]
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Thus x= \[\pm \pi/2\]

Answer 2

Thank you!