Solution of system of trigonometric equations

[Potty susu]

Can anyone please give a detailed stepwise solution how to find the solution of the following equations

cos (x + y/2) cos(y/2)=0
cos (y + x/2) cos(x/2)=0

the answer is (PI,PI) and (PI/3 ,PI/3) but can plz anyone explain how to find it stepwise

 

[Deleted member 4993]

Can anyone please give a detailed stepwise solution how to find the solution of the following equations

cos (x + y/2) cos(y/2)=0
cos (y + x/2) cos(x/2)=0

the answer is (PI,PI) and (PI/3 ,PI/3) but can plz anyone explain how to find it stepwise

Hint: use cos(A+B) = cosA * cosB - sinA * sinB

What are your thoughts?

Please share your work with us ...even if you know it is wrong

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[Potty susu]

Hint: use cos(A+B) = cosA * cosB - sinA * sinB

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

Then what
iam a beginner plz help
i want to know how to solve two trigonometric equations like these
is there any method?

 

[stapel]

Then what
iam a beginner plz help
i want to know how to solve two trigonometric equations like these
is there any method?

One method might be to apply the hint provided earlier. If you are unable even to make a start, then you may need more help than can here be provided. Are you taking a class, or are you attempting self-study? Thank you!

 

[Ishuda]

Then what
iam a beginner plz help
i want to know how to solve two trigonometric equations like these
is there any method?

In many situations, you can break problems into sets and obtain solutions for several sets which cover everything. This is a good example of such a problem as indicated by the answer. Look at the first equation
cos (x + y/2) cos(y/2)=0
Solutions to this mean that either
(1) cos(y/2)=0
or
(2) cos (x + y/2) = 0

Suppose we start with (1). Then
y/2 = \(\displaystyle \pi\)/2 + n \(\displaystyle \pi\)
or
y = \(\displaystyle \pi\) + 2 n \(\displaystyle \pi\).
Now use Subhotosh Khan's hint
cos(A+B) = cosA * cosB - sinA * sinB
in the second equation
cos (y + x/2) cos(x/2) = 0
and continue from there.

Having done (1), move onto to (2).

 

[Potty susu]

In many situations, you can break problems into sets and obtain solutions for several sets which cover everything. This is a good example of such a problem as indicated by the answer. Look at the first equation
cos (x + y/2) cos(y/2)=0
Solutions to this mean that either
(1) cos(y/2)=0
or
(2) cos (x + y/2) = 0

Suppose we start with (1). Then
y/2 = \(\displaystyle \pi\)/2 + n \(\displaystyle \pi\)
or
y = \(\displaystyle \pi\) + 2 n \(\displaystyle \pi\).
Now use Subhotosh Khan's hint
cos(A+B) = cosA * cosB - sinA * sinB
in the second equation
cos (y + x/2) cos(x/2) = 0
and continue from there.

Having done (1), move onto to (2).

Well thank you ishuda for that method
though I would like to tell u that I have found those two answers by this method only... But I did not require Shubhotosh khans hint for that.
i wanted to discuss this question here to know that is there any specific method to solve such system of equations or not.

 

[Deleted member 4993]

Well thank you ishuda for that method
though I would like to tell u that I have found those two answers by this method only... But I did not require Shubhotosh khans hint for that.
i wanted to discuss this question here to know that is there any specific method to solve such system of equations or not.

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