Question

Solve sin

2
(x)=−6cos(x) for all solutions 0≤x<2π x Give your answers accurate to 2 decimal places, as a list separated by commas

Answer 1

To solve the equation sin^2(x) = -6cos(x) for all solutions 0 ≤ x < 2π, substitute cos(x) as y, solve the quadratic equation y^2 + 6y - 1 = 0, and find the corresponding values of x by taking the inverse cosine of y.

Step-by-step explanation:

To solve the equation sin2(x) = -6cos(x) for all solutions 0 ≤ x < 2π, we can rewrite the equation as 1 - cos2(x) = -6cos(x). Now, let's substitute cos(x) as y, so we have 1 - y2 = -6y.

Rewriting the equation further, we get y2 + 6y - 1 = 0. Solving this quadratic equation will give us the values of y. Then, we can find the corresponding values of x by taking the inverse cosine of y.

Using the quadratic formula, we get y = -3 ± √(10). Taking the inverse cosine of these values will give us the solutions for x.

Answer 2

The trigonometric equation sin(2x) = -6cos(x) has no solutions within the interval 0≤x<2π because the value -6 is outside the range of the sine function, which is [-1, 1].

Step-by-step explanation:

To solve the trigonometric equation sin(2x) = -6cos(x) for all solutions within the interval 0≤x<2π, we first rewrite the equation using trigonometric identities. The double angle identity for sine is sin(2x) = 2sin(x)cos(x). We substitute this into the original equation:

2sin(x)cos(x) = -6cos(x)

Dividing both sides by cos(x), provided cos(x) ≠ 0, we get:

2sin(x) = -6

This equation has no solutions because the sine function has a range of [-1, 1] and -6 is outside this range. Therefore, there are no solutions to the given trigonometric equation within the specified interval.

The properties of trigonometric functions and trigonometric identities are crucial in solving trigonometric equations and should be taken into account while solving such problems.