# Math Help - Trig Function Intersection

1. ## Trig Function Intersection

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!

2. ## hi

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?

3. 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?????

4. Originally Posted by skeske1234
How did you get from 2(1-sin^2x)-4sinx+11 to 6cos^2x+11cosx-7?????
No

$u = \sin(x)$

$4 \sin^{2}(x)+7 \sin(x)+6=2 (1-\sin^{2}(x))-4\sin(x)+11$

$4 u^{2}+7 u+6=2 (1-u^{2})-4u+11$

$6u^{2}+11u-7=0$

5. Maybe I should have been more clear, my apologies.