4 sin x + 2 = 0. [0, 2 pi)

chillywilly12a

chillywilly12a

Answered question

2021-09-14

4sinx+2=0.[0,2π)

Answer & Explanation

AGRFTr

AGRFTr

Skilled2021-09-15Added 95 answers

We are given: 4sinx+2=0
Subtract 2 from both sides: 4sinx=2
Divide both sides by 4: sinx=12
Recall that sine is negative in QIII and QIV.

The reference angle is x'=π6 since sin π6=12.

The QIII solution is:

x=π+x'=π+π6=7π6

The QIV solution is:

x=2πx'=2ππ6=11π6

So, the solutions on [0,2π) are:

x=7π6,11π6

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