Question

solve sin^2x+2sinx+1=0

Answer 1

\[\sin^2x + 2 sinx + 1 = (sinx+1)^2\]
\[(sinx+1)^2 = 0\]
\[sinx + 1 = 0\]
\[sinx = -1\]
\[x = \arcsin(-1)\]

Answer 2

x = 3pi/2 + 2 pi k where k is any integer (if i didn't forget my angles lol)

Answer 3

different question.

Answer 4

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