Trig Function Intersection

**skeske1234** · May 8th 2009, 07:19 AM

Give the exact intersection point for the equations f(x)=4sin^2x+7sinx+6 and g(x)=2cos^2x-4sinx+11

Ok, my result is that there is no intersection point because if you put f(x)=g(x) and try to solve for x or the intersection point, the LS f(x) is not possible, so there is none?

OK can someone just clarify for me if I have this right? Explain please thanks!

**Twig** · May 8th 2009, 07:37 AM

hi

$4 sin^{2}(x)+7 sin(x)+6=2 cos^{2}(x)-4sin(x)+11$

Let $u = sin(x) \mbox{ then we get because } cos^{2}(x) = 1-sin^{2}(x) \, \, \; \; 6u^{2}+11u-7=0$

This has the only valid solution $u = \frac{1}{2}$
But $u=sin(x)=\frac{1}{2} \Rightarrow x = \frac{\pi}{6}$ is a solution. Are there more?

**skeske1234** · May 8th 2009, 10:46 AM

Originally Posted by Twig

hi

$4 sin^{2}(x)+7 sin(x)+6=2 cos^{2}(x)-4sin(x)+11$

Let $u = sin(x) \mbox{ then we get because } cos^{2}(x) = 1-sin^{2}(x) \, \, \; \; 6u^{2}+11u-7=0$

This has the only valid solution $u = \frac{1}{2}$
But $u=sin(x)=\frac{1}{2} \Rightarrow x = \frac{\pi}{6}$ is a solution. Are there more?

How did you get from 2(1-sin^2x)-4sinx+11 to 6cos^2x+11cosx-7?????