Find itnersection (1 Viewer)

[Syd123123]

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]

If you squared at this point, you end up with incorrect solutions. Assuming you did, you will get . If you sub into , 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]

If you squared at this point, you end up with incorrect solutions. Assuming you did, you will get . If you sub into , 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]

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