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.

sin² x − 4 sin x − 2 = 0;    [0, 2𝜋)

x =   (smaller value)

x =   (larger value)

Answer 1

sin²(x) − 4sin(x) − 2 = 0

Let w = sin(x) then re-write the equation:

w² - 4w - 2 = 0 This quadratic is not factorable so use the quadratic formula:

w = [-b ± √(b² - 4ac)]/(2a)

w = [4 ± √(16 - 4(1)(-2))]/(2•1)

w = [4 ± √(16 +8)]/2

w = [4 ± √24]/2

w = [4 ± 2√6]/2

w = 2 ± √6

Now, back-substitute:

w = sin(x)

sin(x) = 2 ± √6

At this point, since you are to "approximate the answers", you can switch to decimal approximations for 2 ± √6

Let's break this into problems A (2 + √6) and B (2 - √6):

Problem A:

sin(x) = 2 + √6

sin(x) = 4.44949 but since sin(x) only oscillates between -1 and 1, there is no solution to this.

Problem B:

sin(x) = 2 - √6

sin(x) = - 0.44949

A calculator gives the following value of "x" when using sin^-1(-0.44949): x = -0.466194 (in radians)

You will need to interpret that. 2π radians is all the way around the circle, Coming back by 0.466194 gives and answer of x = 2π - 0.466194 = 5.8170. But there will also be a solution in Q3 as π + 0.466194 or x = 3.6078.

So those are the answer on [0, 2𝜋): x = 3.6078 and 5.8170