How do you solve #\sin x - 2\cos x - 1= 0#?

1 Answer
Nghi N.
Mar 8, 2017

#90^@ + k360^@#
#216^@86 + k360^@#

Explanation:

sin x - 2cos x = 1
Call t that tan t = sin t/(cos t) = 2
Calculator gives #t = 63^@43#.
Substitute 2 by #(sin t/cos t)# in the equation.
#sin x - (sin t/cos t)cos x = 1#
sin x.cos t - sin t.cos x = cos t = 0.447
Use trig identity:
sin (x - t) = sin x.cos t - sin t.cos x. We get:
#sin (x - 63.43) = 0.447#
Calculator and unit circle give 2 solutions:
a. #x - 63.43 = 26^@57# -->
#x = 90^@ + k360^@#
b. #x - 63.43 = 180 - 26.57 = 153^@43# -->
#x = 153.43 + 63.43 = 216^@86 + k360^@#

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