How do I factorise an expression

2〖sen〗^2 x+3 senx+1=0

KonstantinChe [14]

2 sin²(x) + 3 sin(x) + 1 = 0

(2 sin(x) + 1) (sin(x) + 1) = 0

2 sin(x) + 1 = 0 OR sin(x) + 1 = 0

sin(x) = -1/2 OR sin(x) = -1

The first equation gives two solution sets,

x = sin⁻¹(-1/2) + 2nπ = -π/6 + 2nπ

x = π - sin⁻¹(-1/2) + 2nπ = 5π/6 + 2nπ

(where n is any integer), while the second equation gives

x = sin⁻¹(-1) + 2nπ = -π/2 + 2nπ

2 cot(x) sec(x) + 2 sec(x) + cot(x) + 1 = 0

2 sec(x) (cot(x) + 1) + cot(x) + 1 = 0

(2 sec(x) + 1) (cot(x) + 1) = 0

2 sec(x) + 1 = 0 OR cot(x) + 1 = 0

sec(x) = -1/2 OR cot(x) = -1

cos(x) = -2 OR tan(x) = -1

The first equation has no (real) solutions, since -1 ≤ cos(x) ≤ 1 for all (real) x. The second equation gives

x = tan⁻¹(-1) + nπ = -π/4 + nπ

sin(x) cos²(x) = sin(x)

sin(x) cos²(x) - sin(x) = 0

sin(x) (cos²(x) - 1) = 0

sin(x) (-sin²(x)) = 0

sin³(x) = 0

sin(x) = 0

x = sin⁻¹(0) + 2nπ = 2nπ

2 cos²(x) + 2 sin(x) - 12 = 0

2 (1 - sin²(x)) + 2 sin(x) - 12 = 0

-2 sin²(x) + 2 sin(x) - 10 = 0

sin²(x) - sin(x) + 5 = 0

Using the quadratic formula, we get

sin(x) = (1 ± √(1 - 20)) / 2 = (1 ± √(-19)) / 2

but the square root contains a negative number, which means there is no real solution.

2 csc²(x) + cot²(x) - 3 = 0

2 (cot²(x) + 1) + cot²(x) - 3 = 0

3 cot²(x) - 1 = 0

cot²(x) = 1/3

tan²(x) = 3

tan(x) = ± √3

x = tan⁻¹(√3) + nπ OR x = tan⁻¹(-√3) + nπ

x = π/3 + nπ OR x = -π/3 + nπ

7 0

3 years ago

For an integer n ≥ 0, show that (n/2) - (-n/2) = n. Use greatest integer function

mars1129 [50]

The greatest integer function returns the largest integer smaller than the number you provide it. That is, if n ≤ x < n + 1, where n is an integer, then the "greatest integer of x" is [x] = n.

• Let n be even. Then we can write n = 2k for some integer k ≥ 0. Now,

[n/2] = [k] = k

while

[-n/2] = [-k] = -k

so that

[n/2] - [-n/2] = 2k = n

• Let n be odd. Then n = 2k + 1 for some integer k ≥ 0. Every odd integer occurs between two even integers, so that

n - 1 < n < n + 1

or equivalently,

2k < n < 2k + 2

so that

k < n/2 < k + 1

It follows that [n/2] = k.

Similarly, if we negative the previous inequality, we have

-k > -n/2 > -(k + 1), or -k - 1 < -n/2 < -k

which means [-n/2] = -k - 1.

So we make the same conclusion,

[n/2] - [-n/2] = k - (-k - 1) = 2k + 1 = n

3 0

2 years ago

There are 3 consecutive odd integers with a sum of 45. What is the greatest of the 3 integers

jeyben [28]

Answer:

15,17,19

so the answer is 17

3 0

2 years ago

What is the complimentary angle of 72 degrees

Irina18 [472]

Answer: 18°

Step-by-step explanation:

Complementary angles are any two angles whose sum is 90 degrees. Hence, if angles x and y are complimentary, we can say

x + y = 90°

Therefore, the complimentary angle of 72° is obtained by subtracting 72° from 90°

i.e 90° - 72° = 18°

Thus, the complimentary angle of 72° is 18°

4 0

3 years ago

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One side of triangle is 4cm longer than the shortest side and 2cm shorter than the longest side the perimeter is 38cm what are t

o-na [289]

9.3, 13.3, 15.3
Hope it helps!

7 0

3 years ago

Read 2 more answers