How do you find all values of x in the interval [0, 2pi] in the equation #cos^2theta -sin^2theta + sintheta= 0#?

1 Answer
Nghi N.
Feb 17, 2016

#x = pi/2, (7pi)/6 and (11pi)/6#

Explanation:

Replace in the equation #cos^2 x# by #1 - sin^2 x# -->
#1 - sin^2 x - sin^2 x + sin x = 0#
#- 2sin^2 x + sin x + 1 = 0.#
Solve this quadratic equation in sin x.
Since a + b + c = 0, use shortcut. The 2 real roots are: sin x = 1 and #sin x = c/a = -1/2#.
a. sin x = 1 --> arc #x = pi/2, or 90^@#
b. #sin x = -1/2# --> #x = -pi/6, or (11pi)/6# (co-terminal)
and #x = -(5pi)/6#, or #(7pi)/6# (co-terminal)
Answers#pi/2, (7pi)/6 and (11pi)/6#

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