Two much trig :)

solve 2cosX + tanX = secX where X is greater than or equal to zero and less than 2 pi. I understand that the secant is basically the inverse of the cosine, (hyp/adj), but I have no idea how to solve for this. I also don't understand how to solve an equation with 2 different trig functions in it (sin and cos).

2*cos(x) + tan(x) = sec(x)
2*cos(x) + sin(x)/cos(x) = 1/cos(x)
2*cos(x)^2 + sin(x) = 1
Because sin(x)^2 + cos(x)^2 = 1,
2*(1-sin(x)^2) + sin(x) = 1
2 - 2*sin(x)^2 + sin(x) = 1
1 - 2*sin(x)^2 + sin(x) = 0
Let y = sin(x), then
1 - 2y^2 + y = 0
Quadratic. y = 1 or -1/2
sin(x) = 1 or sin(x) = -1/2
x = Pi/2 (in domain) or x = -Pi/6 (out of domain)

Thanks Paul,

How did you use the fact that
sin(x)^2 + cos(x)^2 = 1

I am missing the intermediate step.

Thank you for helping though!

Because sin(x)^2 + cos(x)^2 = 1, cos(x)^2 = 1 - sin(x)^2

Quote:

---

Originally Posted by paultwang

2*cos(x) + tan(x) = sec(x)
2*cos(x) + sin(x)/cos(x) = 1/cos(x)
2*cos(x)^2 + sin(x) = 1
Because sin(x)^2 + cos(x)^2 = 1,
2*(1-sin(x)^2) + sin(x) = 1
2 - 2*sin(x)^2 + sin(x) = 1
1 - 2*sin(x)^2 + sin(x) = 0
Let y = sin(x), then
1 - 2y^2 + y = 0
Quadratic. y = 1 or -1/2
sin(x) = 1 or sin(x) = -1/2
x = Pi/2 (in domain) or x = -Pi/6 (out of domain)

---

Pi/2 also is not in the domain ...as tan(x) and sec(x) are not defined at x=pi/2
hence the equation has no solutions
method of arriving at the solution is correct but.... small error in concluding the answer

-pi/6 was found to be one of the two possible values of angle x. And it was said to be not in the domain of x.

The domain of x was mentioned to be
x >= 0, and x < 2pi
Rewriting that,
0 <= x < 2pi
which means x can be from 0 to almost 2pi. Meaning x can be 0 but x cannot be 2pi.

sin x = -1/2
x = arcsin(-1/2) = -pi/6
that is if we measure x clockwise. But even here -pi/6 is in the 4th quadrant.

What if we measure by the usual way---by counterclockwise? Which is intended by the given domain anyway?
What is x then when sin x = -1/2?

Sine is negative in the 3rd and 4th quadrants.
If sin x = 1/2, then x = pi/6.
How would you place this pi/6 then in the 3rd and 4th quadrants so that you can get sin x = -1/2?

You add pi/6 to pi, and you subtract pi/6 from 2pi.
Or,
x = pi +pi/6 = 7pi/6 ....answer.
x = 2pi -pi/6 = 11pi/6 ....answer.

In degrees,
pi/6 = 30 degrees; pi = 180 deg; and 2pi = 360 deg
So,
x = 180 +30 = 210 deg ....answer.
x = 360 -30 = 330 deg ....answer.

Those two values of x are in the given domain.