kittenfelidae
New member
- Joined
- Apr 26, 2020
- Messages
- 9
Thank you so much Mark. I think I'll manage from here.Hello, and welcome to FMH!
I would begin by equating the two curves, and solve for \(x\):
[MATH]\cos(x)=\tan(x)[/MATH]
Multiply by \(\cos(x)\) to obtain:
[MATH]\cos^2(x)=\sin(x)[/MATH]
Apply a Pythagorean identity to the LHS:
[MATH]1-\sin^2(x)=\sin(x)[/MATH]
At this point, we should recognize that we have a quadratic equation in \(\sin(x)\) and arrange in standard form:
[MATH]\sin^2(x)+\sin(x)-1=0[/MATH]
Can you proceed?