Original post by madhadiya
how would you solve 0<x<360
cos2x=1/3
thanks
cos2x=1/3
thanks
What have you got so far? and what can't you figure out?
C3? Isn't this C2?
Original post by buxtonarmy
C3? Isn't this C2?
yeah, this is core 2. A core 3 question would be something like sec^2(2x) = 25 or something
No, it's C3. You have to use the double angle rule as you have the Cos2X there.
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
If you can't:
Gather your like terms and simplify:
1 - 1/3 = 2Sin^2X --> 2/3 = 2Sin^2X
1/3 = Sin^2X
Then square root since Sin^x is actually (Sinx)^2
sqrt(1/3) = Sin X
X = Sin^-1(sqrt(1/3))
That solves for 35.26 degrees.
Then I find it's easiest to draw the graph of Sin (since you have that last bit of your previous workings as Sin)
And see where you have your other solution.
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
If you can't:
Gather your like terms and simplify:
1 - 1/3 = 2Sin^2X --> 2/3 = 2Sin^2X
1/3 = Sin^2X
Then square root since Sin^x is actually (Sinx)^2
sqrt(1/3) = Sin X
X = Sin^-1(sqrt(1/3))
That solves for 35.26 degrees.
Then I find it's easiest to draw the graph of Sin (since you have that last bit of your previous workings as Sin)
And see where you have your other solution.
(edited 9 years ago)
Original post by Toon800
No, it's C3. You have to use the double angle rule as you have the Cos2X there.
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
Well this kind of question is in core 2 and 3...
You don't need the double angle formula though?
Can't you just make it 2X = cos-1 1/3 and solve from there?
Can't you just make it 2X = cos-1 1/3 and solve from there?
Original post by Toon800
No, it's C3. You have to use the double angle rule as you have the Cos2X there.
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
Don't you think you're over complicating things a bit...
you just multiply your range by 2 so... 0 < 2x < 720
then cos^-1(1/3) = x
find all of the x values by looking at the cos graph from 0 to 720
divide them all by 2
done
Original post by Toon800
No, it's C3. You have to use the double angle rule as you have the Cos2X there.
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
I thought that you could just solve it by doing 2x=cos-11/3 and then when you got the result, divide it by 2 and add/take away 180o... lol
Original post by Imranfr0
I thought that you could just solve it by doing 2x=cos-11/3 and then when you got the result, divide it by 2 and add/take away 180o... lol
Yeah lol that's what I thought??
Original post by buxtonarmy
Yeah lol that's what I thought??
Same, using the double angle formula makes it more complicated than it is xD
Original post by Imranfr0
Same, using the double angle formula makes it more complicated than it is xD
Yep, it's just the C3 way so that's how I solved the equation :P
Original post by Toon800
Yep, it's just the C3 way so that's how I solved the equation :P
I see
Hi, I know this is unrelated to trigonometry but can I have some help on this c3 question please?
The part of the graph between y=e^1/2x, the line y=x and the y axis is rotated about the x axis to make a solid of revolution. Find the
volume of this solid, giving your answer in the form of pi (alnb-c), where a, b and c are integers.
The part of the graph between y=e^1/2x, the line y=x and the y axis is rotated about the x axis to make a solid of revolution. Find the
volume of this solid, giving your answer in the form of pi (alnb-c), where a, b and c are integers.
Hi, I know this is unrelated to trigonometry but can I have some help on this c3 question please?
The part of the graph between y=e^1/2x, the line y=2 and the y axis is rotated about the x axis to make a solid of revolution. Find the
volume of this solid, giving your answer in the form of pi (alnb-c), where a, b and c are integers.
The part of the graph between y=e^1/2x, the line y=2 and the y axis is rotated about the x axis to make a solid of revolution. Find the
volume of this solid, giving your answer in the form of pi (alnb-c), where a, b and c are integers.
Original post by Toon800
No, it's C3. You have to use the double angle rule as you have the Cos2X there.
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
If you can't:
Gather your like terms and simplify:
1 - 1/3 = 2Sin^2X --> 2/3 = 2Sin^2X
1/3 = Sin^2X
Then square root since Sin^x is actually (Sinx)^2
sqrt(1/3) = Sin X
X = Sin^-1(sqrt(1/3))
That solves for 35.26 degrees.
Then I find it's easiest to draw the graph of Sin (since you have that last bit of your previous workings as Sin)
And see where you have your other solution.
Cos2X = 1-2Sin^2 X
So you have 1-2Sin^2 X = 1/3
Can you go from there? :P
If you can't:
Gather your like terms and simplify:
1 - 1/3 = 2Sin^2X --> 2/3 = 2Sin^2X
1/3 = Sin^2X
Then square root since Sin^x is actually (Sinx)^2
sqrt(1/3) = Sin X
X = Sin^-1(sqrt(1/3))
That solves for 35.26 degrees.
Then I find it's easiest to draw the graph of Sin (since you have that last bit of your previous workings as Sin)
And see where you have your other solution.
Thanks
I got up till x=35.3 and x=324.7, but apparently there should be 4 answers??When you square root 1/3 the answer could be positive or negative (I think)
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