Find all solutions to the equation in the interval [0,

iricky827b

iricky827b

Answered question

2021-12-05

Find all solutions to the equation in the interval [0,2π).
sin2xtanx=0

Answer & Explanation

George Burge

George Burge

Beginner2021-12-06Added 16 answers

Use the double-angle identity sin2u=2sinucosu and the quotient identity for tangent:
2sinxcosxsinxcosx=0
Factor out sinx:
sinx(2cosx1cosx)=0
By zero factor property,
sinx=0   2cosx1cosx=0
2cosx=1cosx
cos2x=12
cosx=±12=±22
Recall that on [0,2π),sinx=0 for x=0 and x=π (xaξs),cosx=22 for x=π4 (QI) and x=7π4 (QIV),
and cosx=22 for x=3π4 (QII) and x=5π4 (QIII). So, the solutions in the interval [0,2π) are:
x=0,π4,3π4,π,5π4,7π4
Result:
x=0,π4,3π4,π,5π4,7π4

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