Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places.

Question

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. (Enter your answers as a comma-separated list.)

(cos x)(12 cos x + 4) = 5;     [0, 2𝜋)

x = 

Answer 1

The equation is is not yet in a form that is easily solvable (by hand), because though it is in factored form, which we like, the product of the factors is not = 0, it = 5. So, we distribute the cosx, and subtract 5 from both sides to put the equation in a form that is quadratic (in cosx) on the left and = 0:

12cos²x + 4x - 5 = 0

If you like you can use a dummy variable (not for dummies, just meaning it's a stand-in) to replace cosx with z, though this isn't necessary, it just makes it look more familiar. I won't do that below.

We can also check the discriminant, which = 256, a perfect square, to confirm this is factorable:

(6cosx + 5)(2cosx - 1) = 0 Setting each factor = 0 and solving we get cosx = 1/2 or cosx = - 5/6.

Both values are strictly between -1 and 1, so both equations give 2 solutions on the interval 0 ≤ x < 2π. The first 2 are findable without a calculator, since these are benchmark angles, namely x = π/3 and x = 5π/3.

The other 2 solutions we need to use inverse trig for and a calculator, so find cos^-1(-5/6), which will be an angle in QII. Then find the angle in QIII (since cosx is also - there) that shares a reference angle with cos^-1(-5/6). You can find the 2nd solution by calculating 2π - 1st solution. (Any 2 solutions to a cosine equation will sum to 2π.)

Answer 2

Let u = cos x

Therefore, u * (12u + 4) = 5

12u² + 4u = 5

12u² + 4u - 5 = 0

Solve the quadratic equation.

(2u - 1)(6u + 5) = 0

u = 1/2, -5/6

Lastly substitute the cos x back in for u, so

cos x = 1/2, -5/6

Solve using inverse trig