How would I solve the following trig equation? \sin^2x=1-\cos x

dubinskemab

dubinskemab

Answered question

2022-02-28

How would I solve the following trig equation?
sin2x=1cosx

Answer & Explanation

dinela24k

dinela24k

Beginner2022-03-01Added 7 answers

Recall
sin2x=1cos2x
So you can write the equation as a quadratic, in cosx:
sin2x=1cos(x)1cos2x=1cosx
cos2xcosx=0
cosx(cosx1)=0
Let u=cosx, e.g.
u(u1)=0u=0 or u=1
u=cosx=0x=(2k+1)π2, kZ
u=cosx=1x=2kπ, kZ
Pooja Copeland

Pooja Copeland

Beginner2022-03-02Added 8 answers

We rewrite
sin2(x)=1cos(x)
as
1cos2(x)=1cos(x)
so
cos2(x)=cos(x)
cos2(x)=cos(x)
cos2(x)cos(x)=0
cos(x)(cos(x)1)=0
Now:
cos(x)=0x=π2+kπ, kZ
cos(x)1=0cos(x)=1x=2kπ, kZ

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