Solve the system

Question

Solve for all solutions .

Give all possible answers. Round to two decimal places.



Answer 1

sin²(w) = 6cos(w)

Using the Pythagorean Identity sin²(w) + cos²(w) = 1, we can say sin²(w) = 1 - cos²(w) and we can substitute in "1 - cos²(w)" wherever we see sin²(w) in the original equation giving us:

1 - cos²(w) = 6cos(w) or

cos²(w) + 6cos(w) - 1 = 0

Now, to see how to solve this, let x = cos(w). That gives us:

x² + 6x - 1 = 0

This quadratic is not factor-able so we need to use the quadratic formula to solve it.

x = [-b ± √(b² - 4ac)]/(2a) = [-6 ± √(6² - 4(1)(-1))]/((2)(1)) = (-6 ± √40)/2 = -3 ± √10 so x = 0.162278 or -6.162278

Then, we substitute back cos(w) in place of x so cos(w) = 0.162278 or cos(w) = -6.162278

Since cos(w) must be between -1 and +1, cos(w) = -6.162278 has no solution. To find the solution for cos(w) = 0.162278, we just plug in cos^-1(0.16227) into a calculator to get w = 1.41 radians. This is a solution in Q1 however, we also need to consider the same solution in Q4 or where w = -1.41 radians because that will also provide a cosine of 0.162278. -1.41 radians = 2π - 1.41 = 4.88 radians

So w = 1.41 and 4.88 radians

Answer 2

We are given the equation: sin²(w) = 6cos(w)

Using the trig identity sin²(w) + cos²(w) = 1 and rearranging for sin²(w), we get:

sin²(w) = 1 - cos²(w)

We substitute this into our equation that we we given in order to get:

1 - cos²(w) = 6cos(w)

If we move everything to the right-hand side, we get:

0 = cos²(w) + 6cos(w) - 1

If we let x = cos(w), we then have,

x² + 6w - 1 = 0

You can solve this using either completing the square or the quadratic formula to yield the following values for x:

x₁ = -6.16228 and x₂ = 0.162278

Now if we substitute x = cos(w) back in, we get

cos(w) = -6.16228 and cos(w) = 0.162278

Now, the first solution is an extraneous solution because -6.16 is not contained in the range of cos(w) which is [-1, 1]. As for the second solution, It will have solutions in the first and fourth quadrants since the cosine is positive, so we can do the following

w = cos^-1(0.162278) = 1.41 radians

and for the solution in the fourth quadrant:

w = 2pi - 1.41 = 4.88 radians

Answer 3

Solve sin²(w) = 6cos(w) for all solutions 0≤w<2π.

sin²(w) = 1 - cos²(w) <<trig identity>>, so:

1 - cos²(w) = 6cos(w)

cos²(w) + 6cos(w) = 1

complete the square:

cos²(w) + 6cos(w) +9 = +9 +1

(cos(w) + 3)² = 10

cos(w) + 3 = ±√10

cos(w) = -3 ± 3.162 = -6.162, +0.162

cosine cannot be greater than 1, so cos(w) = 0.162, and w = 1.41 radians

The same cosine will occur at 2pi - w = 6.28 - 1.41 = 4.87 radians

CHECK:

sin²(1.41) = 0.987² = 0.974

6cos(1.41) = 0.162 * 6 = 0.972

Close enough--there's just some rounding error