In the interval #0°<=x<=360°#, how many values of #x# satisfy the equation #3sin^2(x) + sin(x) - 2=0#?

1 Answer
Nghi N.
Apr 7, 2016

#41^@81, 138^@19, 270^@#

Explanation:

Solve this quadratic equation for sin x
#f(x) = 3sin^2 x + sin x - 2 = 0#.
Since a - b + c = 0, use shortcut. The 2 real roots are:
sin x = -1 and #sin x = -c/a = 2/3#.
a. sin x = -1 --> #x = (3pi)/2# or #270^@#
b. #sin x = 2/3 #
Trig unit circle and calculator -->
#sin x = 2/3# --> #x = 41^@81 and x = 180 - 41.81 = 138^@19#
Reminder: sin x = sin (180 - x)
Answers for (0, 360);
#41^281, 138^@19, 270^2#

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