Question

4cos+4sin^2=5 find cos​

Answer 1

Answer:

cos θ = 1/2

Step-by-step explanation:

4 cos θ + 4 sin² θ = 5

Use Pythagorean identity.

4 cos θ + 4 (1 − cos² θ) = 5

4 cos θ + 4 − 4 cos² θ = 5

0 = 4 cos² θ − 4 cos θ + 1

0 = (2 cos θ − 1)²

0 = 2 cos θ − 1

cos θ = 1/2

Answer 2

Answer:

60° and 300°

Step-by-step explanation:

Before beginning to answer this question, you'll need a quick formula that will be very helpful trying to answer questions like this one:

• $sinx^2 + cosx^2 = 1$

You'll see why this formula is useful soon.

Work:

$4cosx+4sinx^2=5$

• The way we want to go about answering this question is very similar to what we've learned in the past: factoring. In order to be able to factor though, we need to have only $cosx$ in our equation instead of both $sinx$ and $cosx$ like the above equation has. this is where the formula I showed earlier comes into play.

• Using the equation I showed earlier, we are going to get rid of the $sinx$.

$sinx^2+cosx^2=1$

• We are going to isolate $sinx^2$ and then substitute what its equal to into the equation.

$sinx^2=-cosx^2+1$

• Now Substitute this back into the original equation.

$4cosx+4(-cosx^2+1)=5$

• Put this into standard form. Then we can factor.

$4cosx-4cosx^2+4=5$

$-4cosx^2+4cosx+4=5$

$4cosx^2-4cosx-4=-5$

$4cosx^2-4cosx+1=0$

• Let $cosx$ be $u$.

$4u^2-4u+1=0$

• Look familiar? We could have factored before just the same, but this ensures there will be fewer mistakes while factoring.

$(2u-1)^2=0$

• Now let's change $u$ back to $cosx$

$(2cosx-1)^2=0$

• Set $2cosx-1$ equal to 0.

$2cosx-1=0$

$2cosx=1$

$cosx=1/2$

• Hopefully if your teacher taught you unit circle properly, you will know that $cosx=1/2$ is asking for an angle. In this case, I know the angle because of memorization, but sadly the only way to learn this stuff is memorization, so I wont be able to show work for how I got my final answer. I'll have to just show calculator-wise how its done.

$(cos^-^1)cosx=1/2(cos^-^1)$

$x=60$ $x=300$