We are given: 4sin⁡x+2=0 Subtract 2 from both sides: 4sin⁡x=−2 Divide both sides by 4: sin⁡x=−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=2π−x′=2π−π6=11π6 So, the solutions on [0,2π) are: x=7π6,11π6

"4sin⁡x+2=0;[0,2π)" was answered by our experts

High School
Geometry
Trigonometry
Trigonometric Functions

Tammy Todd

Answered question

2021-05-21

4 \sin x + 2 = 0; [0, 2 π)

Answer & Explanation

jlo2niT

Skilled2021-05-22Added 96 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:

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

The QIV solution is:

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

So, the solutions on $[0, 2 π)$ are:

$x = 7 \frac{π}{6}, 11 \frac{π}{6}$

Jeffrey Jordon

Expert2021-08-11Added 2605 answers

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