2〖sen〗^2 x+3 senx+1=0 2cotxsecx+ 2secx+cotx+1=0

senx〖cos〗^2 x=senx

2〖cos〗^2 x+2senx-12=0

2〖csc〗^2 x+〖cot〗^2 x-3=0

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

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.  

answer: 34rtyj,jkhtyetwrhyjeuiejnbhdfknbkguxfrbhkfdbjvkmbd,kvbsd.kxjvbhjnfbgjxdfn

step-by-step explanation: