How do you solve the equation on the interval [0,2pi) for #2sin (t) cos (t) + sin (t) -2 cos (t) -1 =0 #?
1 Answer
Jun 10, 2016
Explanation:
Group the terms of the equation in order to factor it:
f(t) = (2sin t.cos t - 2cos t) + (sin t - 1 )= 0
f(t) = 2cos t(sin t - 1) + (sin t - 1) = 0
f(t) = (sin t - 1)(2cos t + 1) = 0
Solve the 2 binomials by using Trig Table and unit circle -->
a. sin t - 1 = 0 --> sin t = 1 -->
b. 2cos t + 1 = 0 -->
Note that the arc
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