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2 Questions, i cant solve
- Thread starter jammy95
- Start date
Let me get you started on the first one.
You need to complete the square in the denominator.
Half of \(\displaystyle -6\) is \(\displaystyle -3\)
\(\displaystyle (-3)^2 = 9\)
So add and subtract \(\displaystyle 9\) to the denominator.
One more step will get you into the appropriate form.
The maximum occurs when the square term in the denominator is zero.
You need to complete the square in the denominator.
Half of \(\displaystyle -6\) is \(\displaystyle -3\)
\(\displaystyle (-3)^2 = 9\)
So add and subtract \(\displaystyle 9\) to the denominator.
One more step will get you into the appropriate form.
The maximum occurs when the square term in the denominator is zero.
2nd listed problem
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If you can tell the difference between the degree symbol and a zero exponent when typed,
then the given problem should be given to you by the instructor and/or the textbook as:
\(\displaystyle 3\cos(2x)^{\circ} \ - \ 9\cos (x)^{\circ} \ = \ 12, \ \ \ in \ the \ interval \ \ 0 \ \le \ x \ \le \ 360\)
Instead, as well as placing grouping symbols around the angle expressions with the degree symbol,
I also recommend the equation be typed more clearly/less cluttered as:
\(\displaystyle 3\cos(2x) \ - \ 9\cos(x) \ = \ 12, \ \ \ in \ the \ interval \ \ 0^{\circ}\ \le \ x \ \le \ 360^{\circ}, \ or \ as:\)
\(\displaystyle 3\cos(2x) \ - \ 9\cos(x) \ = \ 12, \ \ \ in \ the \ interval \ \ 0 \ degrees \ \le \ x \ \le \ 360 \ degrees\)
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\(\displaystyle An \ identity \ (one \ of \ them) \ for \ \cos(2x) \ is \ 2\cos^2(x) - 1.\)
If you substitute this into the equation, you'd have:
\(\displaystyle 3[2\cos^2(x) - 1] \ - \ 9\cos(x) \ = \ 12.\)
This can be altered to:
\(\displaystyle 6\cos^2(x) - 3 \ - \ 9\cos(x) \ = \ 12.\)
\(\displaystyle With \ a \ few \ more \ steps, \ you \ can \ alter \ it \ to:\)
\(\displaystyle 6\cos^2(x) \ - \ 9\cos(x) \ - \ 15 \ = \ 0.\)
You can divide both sides of the equation by the greatest common factor
(or also known as the greatest common divisor), which is 3:
\(\displaystyle 2\cos^2(x) \ - \ 3\cos(x) \ - \ 5 \ = \ 0.\)
This is a quadratic equation in terms of cos(x). Factor it if you can:
\(\displaystyle [2\cos(x) \ - \ 5][\cos(x) \ + \ 1] \ = \ 0.\)
Set each factor equal to 0, and solve for any possible appropriate values
of x that fall in the interval specified in the original problem:
-----------------------
If you can tell the difference between the degree symbol and a zero exponent when typed,
then the given problem should be given to you by the instructor and/or the textbook as:
\(\displaystyle 3\cos(2x)^{\circ} \ - \ 9\cos (x)^{\circ} \ = \ 12, \ \ \ in \ the \ interval \ \ 0 \ \le \ x \ \le \ 360\)
Instead, as well as placing grouping symbols around the angle expressions with the degree symbol,
I also recommend the equation be typed more clearly/less cluttered as:
\(\displaystyle 3\cos(2x) \ - \ 9\cos(x) \ = \ 12, \ \ \ in \ the \ interval \ \ 0^{\circ}\ \le \ x \ \le \ 360^{\circ}, \ or \ as:\)
\(\displaystyle 3\cos(2x) \ - \ 9\cos(x) \ = \ 12, \ \ \ in \ the \ interval \ \ 0 \ degrees \ \le \ x \ \le \ 360 \ degrees\)
---------------------------------------------------------------------------------------------------------------------
\(\displaystyle An \ identity \ (one \ of \ them) \ for \ \cos(2x) \ is \ 2\cos^2(x) - 1.\)
If you substitute this into the equation, you'd have:
\(\displaystyle 3[2\cos^2(x) - 1] \ - \ 9\cos(x) \ = \ 12.\)
This can be altered to:
\(\displaystyle 6\cos^2(x) - 3 \ - \ 9\cos(x) \ = \ 12.\)
\(\displaystyle With \ a \ few \ more \ steps, \ you \ can \ alter \ it \ to:\)
\(\displaystyle 6\cos^2(x) \ - \ 9\cos(x) \ - \ 15 \ = \ 0.\)
You can divide both sides of the equation by the greatest common factor
(or also known as the greatest common divisor), which is 3:
\(\displaystyle 2\cos^2(x) \ - \ 3\cos(x) \ - \ 5 \ = \ 0.\)
This is a quadratic equation in terms of cos(x). Factor it if you can:
\(\displaystyle [2\cos(x) \ - \ 5][\cos(x) \ + \ 1] \ = \ 0.\)
Set each factor equal to 0, and solve for any possible appropriate values
of x that fall in the interval specified in the original problem: