The general solution of |sin⁡x|=cos⁡x is -(A) 2nπ+π4,n∈I(B) 2nπ±π4,n∈I(C) nπ+π4,n∈I(D) None of theseSo what I...

Terrie Lang

Terrie Lang

Answered

2022-01-01

The general solution of |sinx|=cosx is -
(A) 2nπ+π4,nI
(B) 2nπ±π4,nI
(C) nπ+π4,nI
(D) None of these
So what I did was - I made a case for when sinx is greater than 0 and equated it to cosx to get tanx=1 which implies x=π4. The other case was when cosx=sinx. Here, x=3π4. I don't understand how to proceed from here.

Answer & Explanation

Melissa Moore

Melissa Moore

Expert

2022-01-02Added 32 answers

cosx=|sinx|0
If sinx<0,cosx=sinxtanx=1
As x will lie in the 4th quadrant, x=2nππ4
Jeffery Autrey

Jeffery Autrey

Expert

2022-01-03Added 35 answers

Using Weierstrass substitution with t=tanx2
1t2=2|t|
For real t, t2=|t|2
|t|22|t|1=0
|t|=1±2
As |t|0,|t|=2+1=cscπ4+cotπ4==cotπ8=tan(π2π8)
tan2x2=tan2(π2π8)
x2=nπ±(π2π8)
Vasquez

Vasquez

Expert

2022-01-08Added 457 answers

We will solve sin2(x)=cos2(x), which is obtained upon squaring the equation (so we do not have to deal with sign issues).
sin2(x)=cos2(x)cos2(x)sin2(x)=0cos(2x)=0x=π4+πn,3π4+πnx=π4+2πn,5π4+2πn,3π4+2πn,7π4+2πn
Here, we split into cases of 2πn periodicity to take advantage of the properties of sinand cos
Checking these solutions in the original equation |sin(x)|=cos(x), we find that only π4+2πn and 7π4+2πn work. 7π4+2πn is equivalent to π4+2πn, so the answer is (B) ±π4+2πn
You can do a sanity check by drawing a quick graph and seeing if the functions intersect in the general region of the conjectured solution.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?