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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