Find itnersection (1 Viewer)

Syd123123 · Sep 23, 2015

y= arcsin(3x+1)
y = arccosx

I note that -2/3 =< x =< 0
taking sine of both sides, I'll get a quadratic equation giving x = 0 and x = -3/5 as a solution. Both of which, are in the domain. However, when I sketch these two curves, it is only apparent that x = 0 is a solution and x=-3/5 is not. Why is it that algebraically it didn't work?

rand_althor · Sep 23, 2015

$$Did you end up with $\arcsin{(3x+1)}=\arccos{x} \to \arcsin{(3x+1)}=\arcsin{(\sqrt{1-x^2})} \to 3x+1=\sqrt{1-x^2}?$
If you squared at this point, you end up with incorrect solutions. Assuming you did, you will get $x=0,-\frac{3}{5}$ . If you sub $x=-\frac{3}{5}$ into $3x+1=\sqrt{1-x^2}$ , you'll see that LHS =/= RHS. So you have to test solutions at the end to make sure they're correct. You can read more about this here: https://en.wikipedia.org/wiki/Extraneous_and_missing_solutions.

Syd123123 · Sep 23, 2015

rand_althor said:
$$Did you end up with $\arcsin{(3x+1)}=\arccos{x} \to \arcsin{(3x+1)}=\arcsin{(\sqrt{1-x^2})} \to 3x+1=\sqrt{1-x^2}?$
If you squared at this point, you end up with incorrect solutions. Assuming you did, you will get $x=0,-\frac{3}{5}$ . If you sub $x=-\frac{3}{5}$ into $3x+1=\sqrt{1-x^2}$ , you'll see that LHS =/= RHS. So you have to test solutions at the end to make sure they're correct. You can read more about this here: https://en.wikipedia.org/wiki/Extraneous_and_missing_solutions.

Right. Thanks. THat didn't really explain why it doesn't allow for the solution but how to confirm it. But after you wrote your working out I realised that sqrt(1+x^2) = 3x+1 implies 3x+1 > 0 ==> x>-1/3 which means x=-3/5 is not a solution
Also, thanks for the link

braintic · Sep 25, 2015

Alternatively, make siny and cosy the subjects, then square and add.

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