How do you solve $3 sin x + 2 = 0$ $3sinx + 2 = 0$ from 0 to 2pi?

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Answer 1 · 2015-10-28T16:16:04

$x \approx 5.5534576509525865$ $x~~5.5534576509525865$ or $x \approx 3.8713203098167932$ $x~~3.8713203098167932$

$[1] 3 sin x + 2 = 0$ $[1]" "3sinx+2=0$

Subtract 2 from both sides.

$[2] 3 sin x = - 2$ $[2]" "3sinx=-2$

Divide both sides by $3$ $3$ .

$[3] sin x = - \frac{2}{3}$ $[3]" "sinx=-2/3$

Get the inverse sine using a calculator.

$[4] arcsin (- \frac{2}{3}) = x$ $[4]" "arcsin(-2/3)=x$

$[5] arcsin (- \frac{2}{3}) \approx - 0.729727656227$ $[5]" "arcsin(-2/3)~~-0.729727656227$

Since the angle must be from $0$ $0$ to $2 π$ $2pi$ , get an angle that is coterminal with $- 0.729727656227$ $-0.729727656227$ .

$[6] - 0.729727656227 + 2 π \approx 5.5534576509525865$ $[6]" "-0.729727656227+2pi~~5.5534576509525865$

The associate angle of $5.5534576509525865$ $5.5534576509525865$ can be used as well.

$[7] 0.729727656227 + π \approx 3.8713203098167932$ $[7]" "0.729727656227+pi~~3.8713203098167932$

How do you solve 3sinx + 2 = 03sinx+2=03sinx + 2 = 0 from 0 to 2pi?