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1 vote
Which of the following is a solution to 4sin^2x-1=0

a. 30 degrees
b. 60 degrees
c. 90 degrees
d. 120 degrees

User Justanr
by
8.4k points

2 Answers

3 votes


4\sin^2x-1=0\qquad\text{add 1 to both sides}\\\\4\sin^2x=1\qquad\text{divide both sides by 4}\\\\\sin^2x=(1)/(4)\to\sin x=\pm\sqrt{(1)/(4)}\\\\\sin x=-(1)/(2)\ \vee\ \sin x=(1)/(2)\\\\\sin x=-(1)/(2)\to\ x=210^o+360^on\ or\ x=330^o+360^on\qquad n\in\mathbb{Z}\\\\\sin x=(1)/(2)\to x=30^o+360^on\ or\ x=150^o+360^on\qquad n\in\mathbb{Z}\\\\Answer:\ \boxed{a.\ 30\ degrees}

User Wilver
by
8.4k points
3 votes

Hello from MrBillDoesMath!

Answer: a. 30 degrees

Discussion:

To solve 4 (sinx) ^2 - 1 = 0 add 1 to each side:

4 (sinx)^2 = 1. Divide each side by 4:

(sinx)^2 = 1/4 Take the square root of each side

sinx = 1/2


One solution to this equation is x = 30 degrees (choice a)

Thank you,

MrB

User Ben Sharpe
by
7.6k points
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