Sam G.

asked • 07/25/14

How do you solve trig equations?

Solve for x, where 0</=x</=2pi (T)

a. ((1+sinx)/(cosx))+((cosx)/(1+sinx))=4
b. sinxtanx+tanx+cosx = 0

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Olivia B. answered • 07/25/14

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For part b)
 
sin(x)tan(x) + tan(x) + cos(x) = 0
sin(x)[sin(x)/cos(x)] + [sin(x)/cos(x)] + [cos2(x)/cos(x)] = 0
[sin2(x) + cos2(x) + sin(x)]/cos(x) = 0
[1 + sin(x)]/cos(x) = 0
 
This fraction equals zero when the numerator equals zero, i.e., when (1+sin(x))=0, i.e., when sin(x) = -1.
 
The properties of sine tell us that sin(x) = -1 at 3*pi/2 + 2pi*n, for every integer n.
 
Therefore, x = 3*pi/2 + 2pi*n
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Olivia B.

For part a), begin by multiplying through by the product of the denominators, cos(x)*(1+sin(x)):
 
(1+sin(x))2 + cos2(x) = 4cos(x)(1+sin(x))
1 + 2sin(x) + sin2(x) + cos2(x) = 4cos(x)(1+sin(x))
2 + 2sin(x) = 4cos(x)(1+sin(x))
2(1+sin(x)) = 4cos(x)(1+sin(x))
2 = 4cos(x)
1/2 = cos(x)
 
So, it is enough to find x such that cos(x)=1/2.  Properties of cosine tell us that this happens when x = pi/3 + 2pi*n, for every integer n.
 
**Note that there are many different ways of solving these equations.  These are simply two cases.
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07/25/14

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