Respuesta :
Answer:
a) [tex] x= arccos(3/5) = 0.927 [/tex]
Just one solution
b) [tex] cos(x) = -\frac{5}{3}[/tex]
But we know that the cosine can't be negative so then this equation no have solutions on the reals.
c) [tex] x = arcsin(\frac{1}{3})=0.3398[/tex]
This is only the solution on the interval assumed.
d) [tex] x= -0.1145[/tex]
Step-by-step explanation:
a. 5 cos(x) − 3 = 0
For this case we can do this:
[tex] 5 cos(x) = 3[/tex]
Now we can divide both sides by 5 and we got:
[tex] cos (x) = \frac{3}{5}[/tex]
If we apply arccos on both sides we got:
[tex] x = arccos (\frac{3}{5})+2\pi n, x=2\pi-arccos (\frac{3}{5})+2\pi n [/tex]
So for this case the possible solution is:
[tex] x= arccos(3/5) = 0.927 [/tex]
Just one solution. This is only the solution on the interval assumed.
b. 3 cos(x) + 5 = 0
We can do this:
[tex] 3 cos (x) = -5[/tex]
Then we can divide by 3 both sides and we got:
[tex] cos(x) = -\frac{5}{3}[/tex]
But we know that the cosine can't be negative so then this equation no have solutions on the reals.
c. 3 sin(x) − 1 = 0
We can do this:
[tex] 3 sin(x) = 1[/tex]
Then we can divide both sides by 3 and we got:
[tex] sin(x) = \frac{1}{3}[/tex]
The general solutions would be:
[tex] x = arcsin(\frac{1}{3}) + 2\pi n , x= \pi -arcsin (\frac{1}{3}) + 2\pi n[/tex]
[tex] x = arcsin(\frac{1}{3})=0.3398[/tex]
This is only the solution on the interval assumed.
d. tan(x) = −0.115
For this case we can solve the value of x like this:
[tex] x = arctan(-0.115) +\pi n[/tex]
And then the only possible solution for this case is:
[tex] x= -0.1145[/tex]
