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



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


2022-01-02Added 32 answers

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

Jeffery Autrey


2022-01-03Added 35 answers

Using Weierstrass substitution with t=tanx2
For real t, t2=|t|2
As |t|0,|t|=2+1=cscπ4+cotπ4==cotπ8=tan(π2π8)



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).
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.

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