cos(2 x) + 2 = sin(x)
Solve for x over the real numbers:
sin(x) - cos(2 x) = 2
Transform sin(x) - cos(2 x) into a polynomial with respect to sin(x) using cos(2 x) = 1 - 2 sin^2(x):
-1 + sin(x) + 2 sin^2(x) = 2
Divide both sides by 2:
-1/2 + sin(x)/2 + sin^2(x) = 1
Add 1/2 to both sides:
sin(x)/2 + sin^2(x) = 3/2
Add 1/16 to both sides:
1/16 + sin(x)/2 + sin^2(x) = 25/16
Write the left hand side as a square:
(sin(x) + 1/4)^2 = 25/16
Take the square root of both sides:
sin(x) + 1/4 = 5/4 or sin(x) + 1/4 = -5/4
Subtract 1/4 from both sides:
sin(x) = 1 or sin(x) + 1/4 = -5/4
Take the inverse sine of both sides:
x = 2 π n + π/2 for n element Z
or sin(x) + 1/4 = -5/4
Subtract 1/4 from both sides:
x = 2 π n + π/2 for n element Z
or sin(x) = -3/2
sin(x) = -3/2 has no solution since for all x element R, -1<=sin(x)<=1 and -3/2<-1:
Answer: |
| x = 2 π n + π/2 for n element Z
<u>x = 1/2 (4 π n + π)</u> n element Z
Answerep-by-step explanation:
C.10
When adding and subtracting with exponents....combining like terms....exponents are not added/subtracted...but coefficients are.
example : 2b^2 + 3 + 3b^2 = 5b^2 + 3
but keep in mind, they have to be like terms to be able to add and subtract them...for instance 2b^2 + 3b^3...u cannot add/subtract them because they are not like terms.
There are various ways in which you could do this problem. I'm going to share what I think is one of the faster ways.
Instead of thinking of jumping from (-1,4) TO (2,-2), Consider the horiz. jump separately and the vertical jump separately. From -1 to 2 is 3 units. Three times that is 9 units. Add 9 to -1, obtaining 8. That's the horiz. component of the terminal point.
From 4 to -2 is -6 units. Mult. that by 3. The result is the vert. comp of the terminal point.
Note that
sin(x + y) = sin(x) cos(y) + cos(x) sin(y).
Therefore by setting x = π/2 and y = π/7, obtain
sin(π/2 + π/7) = sin(π/2)*cos(π/7) + cos(π/2)*sin(π/7)
The right side is what we want to evaluate. It is equal to
sin(π/2 + π/7) = sin (9/14)π
Answer: