Question

Solve each trigonometric equation such that 0 ≤ x ≤ 2????. Round answers to three decimal places.

Answer 1

Answer:

a) $x= arccos(3/5) = 0.927$

Just one solution

b) $cos(x) = -\frac{5}{3}$

But we know that the cosine can't be negative so then this equation no have solutions on the reals.

c) $x = arcsin(\frac{1}{3})=0.3398$

This is only the solution on the interval assumed.

d) $x= -0.1145$

Step-by-step explanation:

a. 5 cos(x) − 3 = 0

For this case we can do this:

$5 cos(x) = 3$

Now we can divide both sides by 5 and we got:

$cos (x) = \frac{3}{5}$

If we apply arccos on both sides we got:

$x = arccos (\frac{3}{5})+2\pi n, x=2\pi-arccos (\frac{3}{5})+2\pi n$

So for this case the possible solution is:

$x= arccos(3/5) = 0.927$

Just one solution. This is only the solution on the interval assumed.

b. 3 cos(x) + 5 = 0

We can do this:

$3 cos (x) = -5$

Then we can divide by 3 both sides and we got:

$cos(x) = -\frac{5}{3}$

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:

$3 sin(x) = 1$

Then we can divide both sides by 3 and we got:

$sin(x) = \frac{1}{3}$

The general solutions would be:

$x = arcsin(\frac{1}{3}) + 2\pi n , x= \pi -arcsin (\frac{1}{3}) + 2\pi n$

$x = arcsin(\frac{1}{3})=0.3398$

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:

$x = arctan(-0.115) +\pi n$

And then the only possible solution for this case is:

$x= -0.1145$