How do you find #sinx=1/2#?

3 Answers
Apr 14, 2015

Use the trig conversion table and the trig unit circle to solve #sin x = 1/2.#
Trig table gives #sin x = 1/2 = sin (pi/6) --> x_1 = pi/6#.
Trig circle gives another arc #x_2 = 5pi/6# that has the same sin value #(1/2)#.
Since #f(x) = sin x# is a periodic function, with period #2pi#, then there are an infinity of arcs that have the same sin value #(1/2)#, when the variable arc x rotates around the trig unit circle many times. They are called "extended answers".

They are: #x = pi/6 + K*2pi#; and #x = 5pi/6 + K*2pi.# (#K# is a whole number)

Apr 19, 2015

*Diagram not drawn to scale.

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Mar 15, 2017

#30^circ# #and 150^circ#

Explanation:

We can find the answer using a triangle or the unit circle

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Using a right triangle

enter image source here

#color(blue)(sin(theta)=1/2#

As we see in the diagram, #sin(30^circ)# has a opposit and hypotenuse #1 and 2#

So,

#color(green)(rArrsin(30^circ)=1/2#

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Using unit circle

enter image source here

As #30^circ# and #150^circ# has a #sin# of #1/2#,

#color(green)(rArrsin(30^circ)=1/2#

#color(green)(rArrsin(150^circ)=1/2#

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Hope this helps! :)