Solving a trigonometric equation

Question

How can I solve the equation 4 sin²x + 8 cos x - 7 =0 for 0° ≤ x ≤ 360° ?

Answer 1

Wouldn't it be nice if sin x in that equation there was cos x? Then we'd have 4 cos² x + 8 cos x - 7 = 0, and huh, that kind of looks like a quadratic equation which we know how to solve. If we just let u = cos x for a second, then we'd have 4u² + 8u - 7 = 0, which we can solve by the quadratic formula!

In these kinds of problems where it looks almost like a polynomial but not quite, we try to see if we can turn it into a polynomial somehow. One of our best tools is the Pythagorean identity sin² x + cos² x = 1. In case you don't know why this is true, recall that sin x = opposite side of angle x / hypotenuse of the triangle, and cos x = adjacent side of angle x / hypotenuse of the triangle. Let's label opposite side of angle x with a and adjacent side with b. Then we can see that:

sin² x + cos² x = a² / c² + b² / c² = (a² + b²) / c² = c² / c² = 1!

We use the Pythagorean equation here when we substitute a² + b² = c², hence why the identity sin² x + cos² x = 1 is called the Pythagorean trigonometric identity (or just the Pythagorean identity).

So since sin² x + cos² x = 1, we have sin² x = 1 - cos² x. Not exactly what we wanted with just directly making all the terms cos x 's, but we can make it work! Plugging this in, we have

4(1 - cos² x) + 8 cos x - 7 = 0

4 - 4 cos² x + 8 cos x - 7 = 0

-4 cos² x + 8 cos x - 3 = 0 as 4 - 7 = -3.

Alrighty, we're getting somewhere. Now setting u = cos x, we have:

-4u² + 8u - 3 =0

Let's use the quadratic formula. We get

u = [ -8 ± √(64 - 4 • -3 • -4) ] / 2 • -4

u = 1 ± -√(64 - 48) / 8 = 1 ± √16 / 8

u = 1 ± -4 / 8 = 1 ± 1/2

u = 1/2, 3/2

So are our solutions u = 1/2, 3/2? No! Remember that we're solving for x, and cos x = u. So we have x = cos^-1 1/2 and x = cos^-1 3/2 as potential solutions. It's actually relatively easy to remember we're not done when we get to the u = 1/2, 3/2 spot as we know that at least 1/2 is a good number for trig functions. Let's think about when x = cos ^-1 1/2 is true. On the unit circle, the hypotenuse is 1 as the radius is always 1, so since cos = adjacent / hypotenuse, that means that the adjacent side, aka the x-coordinate, is 1/2 when looking at angle x. This happens at x = 60° and 360 - 60º = 60º if you know your unit circle. (Remember that cos -x = cos x, so since we knew cos 60° = 1/2, it must be that cos -60° = 1/2, and cos -60° = cos 360° - 60° = cos 300° as it loops around the circle.

Recall that cos x <= 1. So cos x is never equal to 3/2. Thus our only solutions are 60° and 300°, and the problem is solved!

So to summarize, the strategy is:

1. Convert to a quadratic equation using trig identities
2. Find the solutions of the quadratic equation
3. Figure out what angles create those solutions.

Answer 2

Since cos²x + sin²x = 1, sin²x = 1 - cos²x.

So, we have 4(1 - cos²x) + 8cosx - 7 = 0

4cos²x - 8cosx + 3 = 0

(2cosx - 3)(2cosx - 1) = 0

cosx = 3/2 or cosx = 1/2

Since -1 ≤ cosx ≤ 1 for all real x, the equation cosx = 3/2 has no solution.

cosx = 1/2 when x = π/3 or 5π/3 (in degrees, 60° or 300°)