Answer:
sin(2x)=cos(π2−2x)
So:
cos(π2−2x)=cos(3x)
Now we know that cos(x)=cos(±x) because cosine is an even function. So we see that
(π2−2x)=±3x
i)
π2=5x
x=π10
ii)
π2=−x
x=−π2
Similarly, sin(2x)=sin(2x−2π)=cos(π2−2x−2π)
So we see that
(π2−2x−2π)=±3x
iii)
π2−2π=5x
x=−310π
iv)
π2−2π=−x
x=2π−π2=32π
Finally, we note that the solutions must repeat every 2π because the original functions each repeat every 2π. (The sine function has period π so it has completed exactly two periods over an interval of length 2π. The cosine has period 23π so it has completed exactly three periods over an interval of length 2π. Hence, both functions repeat every 2π2π2π so every solution will repeat every 2π.)
So we get ∀n∈N
i) x=π10+2πn
ii) x=−π2+2πn
iii) x=−310π+2πn
(Note that solution (iv) is redundant since 32π+2πn=−π2+2π(n+1).)
So we conclude that there are really three solutions and then the periodic extensions of those three solutions.
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Answer:
x - 2y = 2
Step-by-step explanation:
the equation of a line in standard form is
Ax + By = C ( A is a positive integer and A, B are integers )
given y = 
x - 1
multiply through by 2 to eliminate the fraction
2y = x - 2 ( subtract 2y from both sides )
0 = x - 2y - 2 ( add 2 to both sides )
x - 2y = 2 ← in standard form
Answer:
they will pay 656 more dollars
Step-by-step explanation:
you multiply 42 by 48 and then subtract it from the full payment
16 weeks. Just add 23 + 4 +4 and so on to get your answer.
Answer:
So option b is right.
Step-by-step explanation:
Given that a figure is located in Quadrant I.
The figure is transformed and the image is located in quadrant iv.
We have to select the option which would have resulted in this.
A) rotating 360 degrees would result in the same place i.e. I quadrant hence wrong.
B) rotating 180 degrees counterclockwise, will make y coordinate negative but x coordinate will remain as it is.
Hence we get image in the IV quadrant.
THis option is correct
C) Rotating90 degrees counterclockwise will not transform all the coordinates to the IV quadrant hence wrong
D) Rotating 270 degrees counterclockwise would make the image in III quadrant hence wrong
So option b is right.
