Question

2. [-/1 Points]

X =
Find the exact solutions of the equation in the interval [0, 2π).
sin 2x + cos x = 0
X =
X =
X =
Need Help?
DETAILS
Submit Answer
LARPRECALCRMRP7 5.5.012.MI.
(smallest value)
(largest value)
Read It
Watch It
Master It

Answer 1

To find the exact solutions of the equation in the interval [0, 2π), we can use the following steps:

• Rewrite the equation as sin 2x = -cos x using the property sin 2x = 2 sin x cos x.
• Divide both sides by cos x to get tan 2x = -1, assuming cos x ≠ 0.
• Use the inverse tangent function to get 2x = tan⁻¹(-1) + kπ, where k is any integer.
• Simplify to get 2x = -π/4 + kπ, or x = -π/8 + kπ/2.
• Substitute different values of k to get the solutions in the interval [0, 2π).
• The solutions are x = π/8, 3π/8, 5π/8, and 7π/8.

Therefore, the exact solutions of the equation in the interval [0, 2π) are:

X = π/8 (smallest value)

X = 3π/8

X = 5π/8

X = 7π/8 (largest value)

Answer 2

`\sf x = (\pi)/(2)` (Smallest value)

`\sf x = (3\pi)/(2)`

`\sf x = (7\pi)/(6)`

`\sf x = (11\pi)/(6)` (Largest value)

Explanation:

We can use the double angle identity for sine to write the equation as follows:

`\sf 2 sin(x) cos(x) + cos(x) = 0`

Factoring out a cos(x), we get:

`\sf cos(x) (2 sin(x) + 1) = 0`

Therefore,

either cos(x) = 0 or

2 sin(x) + 1 = 0.

Solutions where cos(x) = 0:

The solutions to cos(x) = 0 in the interval [0, 2π) are:

`\sf x = (\pi)/(2) , (3\pi)/(2)`

Solutions where 2 sin(x) + 1 = 0:

Subtracting 1 from both sides of the equation 2 sin(x) + 1 = 0, we get:

2 sin(x) = -1

Dividing both sides by 2, we get:

`\sf sin(x) = -(1)/(2)`

The solutions to sin(x) =
`\sf sin(x) = -(1)/(2)` in the interval [0, 2π) are:

`\sf x = (7\pi)/(6) , (11\pi)/(6)`

Therefore, the complete set of solutions to the equation sin 2x + cos x = 0 in the interval [0, 2π) is:

`\sf x = (\pi)/(2), (3\pi)/(2) , (7\pi)/(6) , (11\pi)/(6)`

So, the answer is:

`\sf x = (\pi)/(2)` (Smallest value)

`\sf x = (3\pi)/(2)`

`\sf x = (7\pi)/(6)`

`\sf x = (11\pi)/(6)` (Largest value)