How do you solve #2sin^2 2x+4sin2x=3.759#?

1 Answer
Nghi N.
Aug 30, 2015

Solve #2sin^2 x + 4sin 2x = 3.759#

Ans: x = 22.22 and x = 67.78 deg

Explanation:

Call sin 2x = t. We get a quadratic equation:
#2t^2 + 4t - 3.759 = 0#
#D = d^2 = b^2 - 4ac = 16 + 30.07 = 46.07# --> #d = +- 6.79#
#t = -4/4 +- 6.79/4 = -1 +- 1.70#

a. #t = sin 2x = 0.70# --> #2x = 44.43# -->and
#2x = 180 - 44.43 = 135.57# deg
#2x = 44.43# --> #x = 22.22# deg
#2x = 135.57# --> #x = 67.78# deg
b. t = sin 2x = -2.70 (Rejected)
Check by calculator.
x = 22.22 -> 4sin 2x = 2.80; 2sin^2 2x = 0.98 -> 2.80 + 0.98 = 3.80 OK
x = 67.78; 4sin 2x = 2.80; 2sin^2 2x = 0.98 -> 2.80 + 0.98 = 3.80. OK

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