How do I solve #cos^2x+4=2sinx-3# given 0<x<360?

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Answer 1 · 2018-04-23T20:48:47

No solutions

#cos^2x+4= 2sinx-3#

Use pythagorean identity:
#1-sin^2x= cos^2x#

So:
#(1-sin^2x)+4= 2sinx-3#

#5-sin^2x= 2sinx-3#

Set the expression equal to 0:
#sin^2x+2sinx-8#

Factor:
#(sinx+4)(sinx-2)#

#sinx=4#
#sinx= 2#

Since the arcsin of the anything above #1# or below #-1# is undefined, this equation has no solutions

This equation has no solutions