How do you solve #5sin2x=2cos2x#?

1 Answer
Nghi N.
Aug 25, 2015

Solve #5sin 2x = 2cos 2x#

Ans: #10.90# and #100.90# deg

Explanation:

f(x) = 5sin 2x - 2cos 2x = 0. divide both sides by 5.
#sin 2x - 2/5 cos 2x = 0#
Call #tan a = sin a/cos a = 2/5 = tan 21.80# deg, we get:
sin 2x.cos a - sin a.cos 2x = 0.
Apply the trig identity: sin (a + b) = sin a.cos b - sin b.cos a
sin (2x - a) = sin (2x - 21.80) = 0
2x - 21.80 = 0 and 2x - 21.80 = pi

a. 2x - 21.80 = 0 --> 2x = 21.80 --> #x = 10.9# deg
b. 2x - 21.80 = 180 --> 2x = 201.80 --> #x = 100.90# deg
Check by calculator.
x = 10.9 --> 2x = 21.80 ; 5sin 2x = 1.86 ; 2cos 2x = 1.86
Therefor: 5sin 2x = 2cos 2x = 1.86. OK
x = 100.90 --> 2x = 201.80 ; 5sin 2x = -1.86 ; 2cos 2x = -1.86.
Therefor: 5sin 2x = 2cos 2x = -1.86.

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