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Solve each trigonometric equation such that 0 ≤ x ≤ 2????. Round to three decimal places when necessary.

a. 2cos(x) − 1 = 0
b. 3 sin(x) + 2 = 0

User BClaydon
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1 Answer

7 votes

Answer:

(A)
(\pi )/(3)

(b)
1.768\pi

Explanation:

It is given in the question that
0\leq x\leq 2\pi

(a) We have given equation
2cosx-1=0


cosx=(1)/(2)


x=cos^(-1)(1)/(2)


x=(\pi )/(3)

(b)
3sinx+2=0


sinx=(-2)/(3)


x=sin^(-1)(-0.666)


x=-41.75^(\circ)=360-41.75=318.25^(\circ)=1.768\pi

User Leonaka
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3.1k points