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 did was - I made a case for when sin⁡x is greater than 0 and equated it to cos⁡x to get tan⁡x=1 which implies x=π4. The other case was when cos⁡x=−sin⁡x. Here, x=3π4. I don't understand how to proceed from here.

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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 did was - I made a case for when sin⁡x is greater than 0 and equated it to cos⁡x to get tan⁡x=1 which implies x=π4. The other case was when cos⁡x=−sin⁡x. Here, x=3π4. I don&#039;t understand how to proceed from here.

Melissa Moore · Accepted Answer

cos⁡x=|sin⁡x|≥0   If sin⁡x&amp;lt;0,cos⁡x=−sin⁡x⇔tan⁡x=−1   As x will lie in the 4th quadrant, x=2nπ−π4

Jeffery Autrey · Accepted Answer

Using Weierstrass substitution with t=tan⁡x2  1−t2=2|t|   For real t, t2=|t|2   ⇒|t|2−2|t|−1=0   ⇒|t|=1±2   As |t|≥0,|t|=2+1=csc⁡π4+cot⁡π4=…=cot⁡π8=tan⁡(π2−π8)   ⇔tan2x2=tan2(π2−π8)   ⇒x2=nπ±(π2−π8)

Vasquez · Accepted Answer

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.

The general solution of |\sin x|=\cos x is - (A)

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