Question #44e15

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Answer 1 · 2017-11-22T05:01:50

Given:

#2cos^2(x) - sin(x) = -1#

Substitute #1 - sin^2(x)# for #cos^2(x)#:

#2(1 - sin^2(x)) - sin(x) = -1#

Distribute the 2:

#2 - 2sin^2(x) - sin(x) = -1#

Combine like terms:

# - 2sin^2(x) - sin(x)+3 =0#

Multiply both sides by -1:

#2sin^2(x) + sin(x)-3 =0#

Factor:

#(2sin(x) + 3)(sin(x)-1) = 0#

#sin(x) = -3/2# and #sin(x) = 1#

We must discard the first root because it is outside the range of the sine function, #-1 <= sin(x) <= 1#.

This leaves only the second root:

#sin(x) = 1#

#x = sin^-1(1)#

#x = pi/2#

Add a term that says that sine function repeats at integer multiples of #2pi#:

#x = pi/2 + 2pin; n in ZZ#