How do you solve #2sinx + 1 =0#?

1 Answer
Acquaintance · Stefan V.
May 27, 2016

#x = (11pi)/6, (7pi)/6#

Explanation:

To solve this equation, go about it as you would any other equation. Get the sin x all by itself.

#2 sin x +1 = 0#

#2 sin x = -1#

#sin x = -1/2#

Then, use the unit circle to find all radian values which have a y-coordinate of #-1/2#, since sin is the #y# value (as opposed to cos, which is the x value).

http://www.math.toronto.edu/preparing-for-calculus/8_trigonometry/we_3_unit_circle.html

As you can see, the coordinates #(-sqrt(3)/2#, #-1/2)# and #(sqrt(3)/2, -1/2)# have #y# values (or #sin# values) of #-1/2#.

The radian correspondents of these coordinates are #(7pi)/6# and #(11pi)/6#, and those are your two answers.

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