HSC Multiple choice question (1 Viewer)

drainbammage · Jul 5, 2015

http://puu.sh/iNUCv/bf0be222e0.png

Why is the answer 2 solutions? Shouldn't (Sinx-1) have one solution and (tanx+2) have 2?

InteGrand · Jul 5, 2015

drainbammage said:
http://puu.sh/iNUCv/bf0be222e0.png

Why is the answer 2 solutions? Shouldn't (Sinx-1) have one solution and (tanx+2) have 2?

The trick is that the $\sin x -1 =0$ solution is at a point where the tan function is undefined (namely $x=\frac{\pi}{2}$ ), and hence the L.H.S. is undefined, so it doesn't count as a solution to the equation.

drainbammage · Jul 5, 2015

InteGrand said:
The trick is that the $\sin x -1 =0$ solution is at a point where the tan function is undefined (namely $x=\frac{\pi}{2}$ ), and hence the L.H.S. is undefined, so it doesn't count as a solution to the equation.

Ohhh now it makes sense, thanks,

HSC Multiple choice question (1 Viewer)

drainbammage

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InteGrand

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