Answer:
bsdjsds
Step-by-step explanation:
If d=6, there are two real solutions.
Rule if d=discriminant:
d>0, two real solutions
d=0, one real solution
d<0, no real solutions (but there are two imaginary solutions)
3. First factor as a difference of cubes:
For the remaining group, apply the double angle identity.
5. seems rather tricky. You might want to post another question for that problem alone...
6. Factorize the left side as a sum of cubes:
From here we have to prove that
We can write everything in terms of sine:
(double angle identity)
(angle sum identity)
After some simplifying, we're left with showing that
or
This last equality follows from what you could the triple angle identity for sine,
<span>f(x) = 2x + 3
</span><span>g(x) = x²-7
</span>(f+g)(x) = f(x) + g(x) = 2x + 3 + x² - 7 = x² + 2x - 4
(f+g)(x) = x² + 2x - 4