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HOW DO YOU FIND ALL VALUES OF X SUCH THAT SIN2X=SINX AND 0≤X≤2Π?

Trigonometry Trigonometric Identities and Equations Solving Trigonometric
Equations

1 ANSWER

Shwetank Mauria
May 22, 2016

x=nπ or x=2nπ±π3

EXPLANATION:

As sin2x=sinx, we have

2sinxcosx=sinx or

2sinxcosx−sinx=0

sinx(2cosx−1)=0

i.e. either sinx=0 which implies x=nπ

or 2cosx−1=0 i.e. cosx=12=cos(±π3)

which implies x=2nπ±π3

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