Sin 2x - sin x=0
Using the trigonometric identity: sin 2x=2 sinx cosx
2 sinx cosx - sinx =0
Common factor sinx
sinx ( 2 cosx -1)=0
Two options:
1) sinx=0
on the interval [0,2π), the sinx=0 for x=0 and x=<span>π=3.1416→x=3.14
2) 2 cosx - 1=0
Solving for cosx
2 cosx-1+1=0+1
2 cosx = 1
Dividing by 2 both sides of the equation:
(2 cosx)/2=1/2
cosx=1/2
cosx is positive in first and fourth quadrant:
First quadrant cosx=1/2→x=cos^(-1) (1/2)→x=</span><span>π/3=3.1416/3→x=1.05
Fourth quadrant: x=</span>2π-π/3=(6π-π)/3→x=5<span>π/3=5(3.1416)/3→x=5.24
Answer: Solutions: x=0, 1.05, 3.14, and 5.24</span>
Given:
Amplitude = 2
Midline = 5
Period of
.
To find:
The cosine function.
Solution:
The general cosine function is:
...(i)
Where, |A| is amplitude,
is period,
is phase shift and D is the mid line.
Period of function is
. So,



Substituting
in (i), we get


Therefore, the required cosine function is
.
A great earthquake would be 100 times stronger than a large one.
Answer:
d
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that,
y = x + 6 equation 1
3x - 3y = -18. Equation 2
Substitute equation 1 into equation 2
3x-3(x+6) = -18
3x - 3x - 18 = -18
-18 = -18
Therefore, it has no solution. Check attachment for echelon matrix solution
B. y = -2x + 5 equation 1
2x + y = -7. Equation 2
Substitute equation 1 into 2
2x -2x + 5 = -7
5 = -7
Since the equation is not balance, then, it has infinite numbers of solution.
C. y = -4x + 11. Equation 1
6x + y = 11. Equation 2
Substitute equation 1 into 2
6x - 4x + 11 = 11
2x = 11 - 11
2x = 0
x = 0
Also, y = -4x + 11
y = 0 + 11 = 11
Then, this has a unique solution I.e. one solution.