How to find solutions to the following equations in the interval 0 ≤ x < 360?

Find solutions to the following equations in the interval 0 ≤ x < 360◦. Round
answers to the nearest 10th degree.
3 sin x − 2 = 0

1 Answer
maganbhai P.
Jun 2, 2018

x~~40^circ or x~~140^circx≈40∘orx≈140∘

Explanation:

Here,

3sinx-2=0 ,where, x in[0^circ,360^circ)3sinx−2=0,where,x∈[0∘,360∘)

=>3sinx=2⇒3sinx=2

=>sinx=2/3 > 0=>I^(st)Quadrant or II^(nd)Quadrant ⇒sinx=23>0⇒IstQuadrantorIIndQuadrant

(i) I^(st)Quadrant :(i)IstQuadrant:

sinx=2/3=>x=sin^-1(2/3)sinx=23⇒x=sin−1(23)

=>x~~(41.81)^circ=>x~~40^circ⇒x≈(41.81)∘⇒x≈40∘

(ii)II^(nd)Quadrant :(ii)IIndQuadrant:

sinx=2/3=>x=180^circ-sin^-1(2/3)sinx=23⇒x=180∘−sin−1(23)

=>x~~180^circ-(41.81)^circ⇒x≈180∘−(41.81)∘

=>x~~(138.19)^circ⇒x≈(138.19)∘

=>x~~140^circ⇒x≈140∘

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