Cos(x+90°) = 2 sin^2 x -pi/2 < x < pi/2 Solve algebraically?

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Answer 1 · 2018-03-13T15:58:22

#x = 0 and x = (11pi)/6#

Use the formula #cos(A + B) = cosAcosB - sinAsinB#.

#cosxcos90˚ - sinxsin90˚ = 2sin^2x#

#-sinx = 2sin^2x#

#0 = 2sin^2x + sinx#

#0 = sinx(2sinx+ 1)#

#sinx= 0 or sinx = -1/2#

We have to respect the given domain of #-pi/2 <x < pi/2#, so

#x = 0, (11pi)/6#

The graph confirms:

enter image source here

Hopefully this helps!