Question 937255: Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
Found 3 solutions by lwsshak3, Alan3354, ikleyn:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
(sinx+cosx)^2=sin^2(x)+2sinxcosx+cos^2(x)=1+sin(2x)
1+sin(2x)=1
sin2x=0
2x=0
x=0
no solution: x not in given domain
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
sin(x) + cos(x) = 1
By inspection, x = pi/2
------
sin(x) + cos(x) = -1
x = pi
Answer by ikleyn(53937)  (Show Source):
You can put this solution on YOUR website! .
Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
~~~~~~~~~~~~~~~~~~~~~~~~
In his post, @lwsshak3 came to conclusion "no solution: x not in given interval."
This answer is INCORRECT. Given equation has 3 (three) solutions in the given interval.
They are  ,  and  .
The reasoning in the post by @l2sshak3 is conceptually wrong,
it is why he missed 3 solutions.
It is wrong way to teach students, so I came to bring a correct solution.
Your starting equation is
 = 1
Transform it step by step
 = 1,
1 + 2*sin(x)*cos(x) = 1,
sin(2x) = 0.
Hence, 2x should be a multiple of .
Since x should be in interval ( , ), 2x should be in interval (, )
0 < 2x < .
There are 3 possible values for 2x: , and  .
It gives 3 possible values for x: , and .
It is easy to check that all these 3 values satisfy given equation.
ANSWER. Given equation has 3 solutions in the given interval: , and .
Solved correctly to teach you in a right way.
|
|
|
