If 2cos²(π+x)+3sin(π+x) vanishes then the values of x lying in the interval from 0 to 2π is x=π/6 or 5π/6. Explain how?

1 Answer
Nghi N.
Jun 21, 2018

We have to prove that #(pi/6)# and #(5pi)/6# are the 2 roots of the equation:
#2cos^2 (pi + x) + 2sin (pi + x) = 0#
First, replace in the equation x by #pi/6, or x = 30^@#, then, evaluate the equation.
#2cos^2 (pi + pi/6) = 2cos^2 ((7pi)/6) = 2cos^2 (210)) = 1.50#
#3sin ((7pi)/6) = 3sin (210) = 3(-0.5) = - 1.50#. Proved.
Next, replace x by #(5pi)/6, or x = 150^@#:
#2cos^2 ((11pi)/6) = 2cos^2 (330) = 2(0.75) = 1.50#
#3sin ((11pi)/6) = 3sin (330) = 3(-0.5) = - 1.50#. Proved

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