7sin x + 5 = 2cos^2 x.7sinx+5=2cos2x.
Replace cos^2 xcos2x by (1 - sin^2 x)(1−sin2x):
7sin x + 5 = 2 - 2sin^2 x7sinx+5=2−2sin2x
2sin^2 x + 7sin x + 3 = 02sin2x+7sinx+3=0
Solve this quadratic equation for sin x.
D = d^2 = b^2 - 4ac = 49 - 24 = 25D=d2=b2−4ac=49−24=25 --> d = +- 5d=±5
There are 2 real roots:
x = -b/(2a) = -7/4 +- 5/4 = (-7 +- 5)/4x=−b2a=−74±54=−7±54
x1 = -2/4 = -1/2x1=−24=−12, and x2 = -12/4 = -3x2=−124=−3 (Rejected because < -1)
Trig table, and unit circle give -->
x = - 1/2x=−12 --> 2 arcs x
a. x = - pi/6x=−π6, or x = (11pi)/6 x=11π6--> (co-terminal), and
b. x = pi + pi/6 = (7pi)/6x=π+π6=7π6
Answers for (0, 2pi)(0,2π)
(7pi)/6 and (11pi)/67π6and11π6
