We are given: 4sin⁡x+2=0Subtract 2 from both sides: 4sin⁡x=−2Divide both sides by 4: sin⁡x=−12Recall 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=7π6The QIV solution is:x=2π−x'=2π−π6=11π6So, the solutions on [0,2π) are:x=7π6,11π6

Get the answer to "4sin⁡x+2=0.[0,2π)"

High School
Geometry
Trigonometry
Trigonometric Functions

chillywilly12a

Answered question

2021-09-14

4 \sin x + 2 = 0 . [0, 2 π)

Answer & Explanation

AGRFTr

Skilled2021-09-15Added 95 answers

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

The reference angle is $x' = \frac{π}{6}$ since sin $\frac{π}{6} = \frac{1}{2}$ .

The QIII solution is:

$x = π + x' = π + \frac{π}{6} = \frac{7 π}{6}$

The QIV solution is:

$x = 2 π - x' = 2 π - \frac{π}{6} = \frac{11 π}{6}$

So, the solutions on $[0, 2 π)$ are:

$x = \frac{7 π}{6}, \frac{11 π}{6}$

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