SOLUTION: Find all solutions of each of the equations in the interval [0,2pi). a) sin(x+pi/3)+sin(x−pi/3)=1 b) tan(x+pi)+2sin(x+pi)=0 c) cos(x−pi/2)+sin2x=0

Algebra ->  Trigonometry-basics -> SOLUTION: Find all solutions of each of the equations in the interval [0,2pi). a) sin(x+pi/3)+sin(x−pi/3)=1 b) tan(x+pi)+2sin(x+pi)=0 c) cos(x−pi/2)+sin2x=0      Log On


   

Question 1077821: Find all solutions of each of the equations in the interval [0,2pi).
a) sin(x+pi/3)+sin(x−pi/3)=1

b) tan(x+pi)+2sin(x+pi)=0

c) cos(x−pi/2)+sin2x=0
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
a) sin%28x%2Bpi%2F3%29%2Bsin%28x-pi%2F3%29=1
Using the trigonometric identities
for sine of sum an difference of two angles,
the equation can be re-written as
sin%28x%29cos%28pi%2F3%29%2Bcos%28x%29sin%28pi%2F3%29%22%2B%22sin%28x%29cos%28pi%2F3%29-cos%28x%29sin%28pi%2F3%29=1
Taking out sin%28x%29 and sin%28x%29 as common factors
the equation can be re-written as
sin%28x%29%28cos%28pi%2F3%29%2Bcos%28pi%2F3%29%29%22%2B%22cos%28x%29%28sin%28pi%2F3%29-sin%28pi%2F3%29%29=1
sin%28x%29%28cos%28pi%2F3%29%2Bcos%28pi%2F3%29%29%22%2B%22cos%28x%29%280%29=1
sin%28x%29%28cos%28pi%2F3%29%2Bcos%28pi%2F3%29%29=1
We know that cos%28pi%2F3%29=1%2F2 , so we re-write the equation as
sin%28x%29%281%2F2%2B1%2F2%29=1 and sin%28x%29=1 .
In the interval [0,2pi), that happens only for
x=pi%2F2 .


b) tan%28x%2Bpi%29%2B2sin%28x%2Bpi%29=0
Based on trigonometric identities, the equation can be re-written as
tan%28x%29-2sin%28x%29=0 and sin%28x%29%2Fcos%28x%29-2sin%28x%29=0 .
Then, with some algebra, it can be re-written as
sin%28x%29%281%2Fcos%28x%29-2%29=0 and sin%28x%29%281-2cos%28x%29%29%2Fcos%28x%29=0
The numerator is zero when
sin%28x%29=0 ---> highlight%28x=0%29 or highlight%28x=pi%29 .
The numerator is also zero when
1-2cos%28x%29=0 ---> cos%28x%29=1%2F2 ---> highlight%28x=pi%2F3%29 or highlight%28x=5pi%2F3%29 .
For none of those values of x, is the cos%28x%29 zero,
so they are all valid solutions.


c) cos%28x-pi%2F2%29%2Bsin%282x%29=0
(Or did you mean cos%28x-pi%2F2%29%2Bsin%5E2%28x%29=0 instead?)
Using trigonometric identities,
the equation can be re-written as
cos%28-%28x-pi%2F2%29%29%2Bsin%282x%29=0 <--> cos%28pi%2F2-x%29%2Bsin%282x%29=0 and sin%28x%29%2Bsin%282x%29=0 .
If the second term was really ,
using the trig identity for double angles,
the equation can be re-written as
sin%28x%29%2B2sin%28x%29cos%28x%29=0 <---> sin%28x%29%281%2B2cos%28x%29%29=0
The expression sin%28x%29%281%2B2cos%28x%29%29 is zero when
sin%28x%29=0 ---> highlight%28x=0%29 or highlight%28x=pi%29 .
The expression sin%28x%29%281%2B2cos%28x%29%29 is also zero when
1%2B2cos%28x%29=0 <---> cos%28x%29=-1%2F2 .
In the interval [0,2pi), that happens for
highlight%28x=2pi%2F3%29 or highlight%28x=4pi%2F3%29 .


NOTE: For cos%28x-pi%2F2%29%2Bsin%5E2%28x%29=0 ,
sin%28x%29%2Bsin%5E2%28x%29=0 <--> sin%28x%29%281%2Bsin%28x%29%29=0 ,
in the interval [0,2pi) has solutions when
sin%28x%29=0 --> x=0 or x=pi ,
and when
1%2Bsin%28x%29=0 --> sin%28x%29=-1 --> x=3pi%2F2 .