How do you solve #sinx = cos2x - 1#?

2 Answers
Zor Shekhtman
Jul 14, 2015

Solutions are:
#x=pi*N#
#x=-pi/6+2pi*N#
#x=pi+pi/6+2pi*N#
where #N# - is any integer number.

Explanation:

First of all, let's transform our equation into a one that depends only on #sin(x)#. For this, recall the trigonometric identity:
#cos(2x) = cos^2(x)-sin^2(x) = 1-2sin^2(x)#

The our equation looks like
#sin(x)=1-2sin^2(x)-1# or
#2sin^2(x)+sin(x)=0# or
#sin(x)*(2sin(x)+1)=0#

From the last equation we have two directions:
(a) #sin(x)=0#, from which a solution #x=pi*N# follows, where #N# - is any integer number.
(b) #2sin(x)+1=0#, from which we can derive #sin(x)=-1/2#, solutions of which are #x=-pi/6+2pi*N# and #x=pi+pi/6+2pi*N#

Nghi N.
Jul 14, 2015

Solve sin x = cos 2x - 1

Explanation:

Replace #cos 2x# by #(1 - 2sin^2 x):#
#sin x = 1 - 2sin^2 x - 1 -> 2sin^2 x + sin x = 0#
#sin x(2sin x + 1) = 0#

a. #sin x = 0 --> x = 0 and x = pi#

b. #sin x = - 1/2 --> x = (7pi)/6# and #x = (11pi)/6#

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