Trigonometry
The function f is defined for all real numbers by f(x)=sin(2x+ π/4) Find the general solution of the equation f(x)=0.
I seriously need help with this!!!
Answers
Hi there. The question tells us that sin(2x+pi/4) = 0, as both are equal to f(x). We use the inverse sin of 0 to tell us that 2x+pi/4 = 0 (the inverse sin of 0 is 0). We can just rearrange this to give. x = -pi/8. Hope this helps
31 December 2013
first, what is inside the ( ) is the "angle"then, know that angles for functions and graphs are measured in radians:0 degrees = 0 radians90 deg = pi/2 rad180 deg = pi rad360 deg = 2pi rad and so onThen, sin(angle) will be 0, everytime that (angle)= 0 or 180 or 360 and so on ( half circles and whole circles)But remember you work in radians not degrees.So sin(2x + pi/4) = 0 when (2x + pi/4) = 0, or pi, or 2pi, or 3pi . . . ... or you could write and solve, in GENERAL, (2x + pi/4) = 0 + (n times)piwhich you can do :)Solution is x=(n - 1/4) pi/2 or (n/2 - 1/8) pi
31 December 2013
Let u=2x+pi/4Then sinu=0 means that u = 0, pi or 2pi etc (ie multiples of pi, incl. -pi etc) (*)So 2x+pi/4 = 0, pi or 2pi etcHence x = -pi/8, 3pi/8 or 7pi/8 etcIf you are told that eg 0<x<pi, then pi/4<2x+pi/4<2pi+pi/4, ie pi/4<u<2pi+pi/4,so only consider u=pi & 2pi in (*) (ie no need to consider u=0 or 3pi etc)Nick
02 January 2014
Sorry, ought to have added that a general expression for -pi/8, 3pi/8 or 7pi/8 etc is (4n-1)pi/8 (as the general formula for the sequence -1, 3, 7 etc is 4n-1)Nick
02 January 2014
f(x)=sin(2x+pi/4), find for f(X)=0. This means sin(2x+pi/4)=0; 2x+pi/4=0 or pi or 2pi and so on in multiples of 2 so next will be 4pi..... This is multiples because sin of 0 or pi or 2pi or 4 pi gives zero. Hence if boundaries are not given general solution would be x= (n/2-1/8)pi where n is 0,1,2,3.... as many as you want.If gcse level maths then x=-pi/8=-0.393 (using calculator)
10 January 2014
y= 2x +pi/4sin(y)= 0 y= 2*k*pi 2x + pi/4 = 2*k*pi x= (2*k*pi - pi/4)/2
10 January 2014
Okay, so for f(x) to =0, the content of the bracket (2x+pi/4) must either =0, as sin(0) is 0, or a multiple of pi. If you look at a sine graph you'll see that the sine of pi or it's multiples is 0.So, from this we can say that 2x+pi/4=0 (or pi, 2pi etc)for the 0 example x would equal -pi/8, here's why:2x+pi/4=02x=-pi/4x=-pi/8This would most likely be your answer, especially if you are given a bounded region. Potentially there could be many answers, but if this is for GCSE it's likely -pi/8.
12 January 2014
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13 February 2014
See image below for graphical solution to problem - look at where the function crosses the x-axis.
13 February 2014
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13 February 2014
Zoomed in version of graphical solution.
13 February 2014
AS F(X)=0:. sin(2x+pi/4)=0 and we know that sin is zero at 0,180,360,...degrees i.e at kpi:. sin(2x+pi/4)=sin(kpi):. 2x+pi/4=kpi:. 2x=kpi+pi/4:.x=((4k+1)pi)/8so the correct ans is x=((4k+1)pi)/8 where k is a integer
11 May 2014
Okay, we know sin(2x+pi/4)=0. Keep in mind that sin("value)"=0 only when the "value" is of the form 0+pi*K, for any integer K. For example, sin(0)=sin(pi)=sin(2*pi)=sin(3*pi)=0, and so on. So, then since the "value"=2x+pi/4, we know that:K*pi=2x+pi/4 for any integer K.Solving for x, we get x=((K-1/4)*pi)/2. for any integer K!And...didn't even need pictures. :)
11 May 2014
as f(x)=0:. sin(2x+pi/4)=0 and we know sin is zero at 0,180,360,.... i.e kpi where k is an integer;. sin(2x+pi/4)=sin(kpi):. 2x+pi/4=kpi;.2x=kpi-pi/4=> x=((4k-1)pi)/8 where k is an integer is the general sol
11 May 2014
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14 May 2014
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14 May 2014
since f(x)= 0sin(2x+pi/4) =0; (2x+pi/4)= npi; 2x= npi-pi/4; so x= npi/2- pi/8;
15 May 2014
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16 May 2014
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04 July 2014
f(x) = sin(2x+pi/4) also f(x) =0 so we can saysin(2x+pi/4) = 0take the inverse sin of both sides to get(ensure calculator is in radians2x+pi/4 = n*pi.subtract pi/4 from both sides to get2x = (n*pi) - (pi/4)divide both sides by 2x = (n*pi/2) - (pi/8)
04 July 2014
f(x)=0sin(2x+pi/4)=02x+pi/4=n(pi)x=1/2(n(pi)-(pi)/4)If you need more help in A level you may contact me....
21 July 2014
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21 July 2014
Hello,I have attached the solution of the mentioned problem. Hope it will help you.all the best.
21 July 2014
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