0
tanx=2cscx
sinx/cosx=2/sinx
sin2x/cosx=2
sin2x=2cosx
1-cos2x=2cosx
0=cos2x+2cosx-1
cosx=(-2±√(2^2+4))/2
cosx=(-2±√8)/2
cosx=(-2±2√2)/2
cosx=-1±√2
cosx=approximately -2.41 or approximately 0.41.
Since the range of the cosine function is [-1,1], only approx. 0.41 works.
So:
cosx= approx. 0.41
Need calculator now (I went as far as I could without one!)
x=approx 1.148
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Intergrate sec x?
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
How do you integrate sec cube 2x?
The integral of sec(x) is ln|secx+tanx| + C Since the derivative is taken to the third power, we have to consider the chain rule; the original equation must be to the fourth power, and in order for that to be canceled out, the equation must also have had a coefficient of 1/4. 2x is also subject to the chain rule. I would suggest u substitution. integral(sec(2x))^3 dx u=2x du=2dx dx=1/2du integral (sec(u))^3 *1/2 du 1/8 secxtanx + 1/8(ln|secx+tanx|^4) + C
Differential co-efficient of log tan x is?
y = ln(tan(x)) u = tanx y =ln(u) dy/du = 1/u du/dx = sec2(x) dy/dx = dy/du * du/dx = sec2(x)/tan(x)
Evaluate Limit x tends to 0 x minus sinx divided by tanx minus x?
lim (x→0) [(x - sin x)/(tan x - x)]Since both the numerator and the denominator have limit zero as x tends to 0, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply the l'Hopital's Rule and the limit equalslim (x→0) [(x - sin x)'/(tan x - x)']= lim (x→0) [(1 - cos x)/(sec2 x - 1)] (form 0/0, use again the l'Hopital's Rule)= lim (x→0) [(1 - cos x)'/(sec2 x - 1)']= lim (x→0) [(0 - (-sin x)/(2sec x sec x tan x - 0)]= lim (x→0) [(sin x)/(2sec2 x tan x)] (substitute 1/cos2 x for sec2 x and sin x/cos x for tan x)= lim (x→0) [(sin x)/(2sin x/cos3 x)]= lim (x→0) [(sin x cos3 x)/2sin x]= lim (x→0) (cos3 x/2)= 1/2Thus, (x - sin x)/(tan x - x) tends to 0.5 as x tends to 0.
Tanx equals 45 is 49 degrees?
No.
Tan plus cot divided by tan equals csc squared?
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
Tan3x plus tanx equals 2tan2x?
Tan
What is period of y equals -tanx?
pi radians.
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sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
What is the differentiate of y equals tanx plus c?
y' = (sec(x))^2
How do you prove the following identity sec x - cos x equals sin x tan x?
you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx
Is there a graph of y equals equals 2 tan 2x?
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How does secx plus 1 divided by cotx equal 1 plus sinx divided by cosx?
secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = 1 + sinx/cosx, andsin/cos = tanx, therefore1/cosx + tanx = 1 + tanx, therefore1/cosx = 1, therfore1 = cosx.So, therfore, it is not neccesarily true.But if you meansecx plus 1 divided by cotx equals (1 plus sinx) divided by cosx(this is probably what you mean) Let's start over!secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = (1+sinx)/cosx therefore1/cosx + tanx = 1/cosx + sinx/cosxsinx/cosx = tanx therfore1/cosx + tanx = 1/cosx + tanxDo you think this is correct? Subtract both sides by 1/cosx + tanx:0 = 0So, therefore, this is correct!(BTW, I'm in Grade 6! :P)
When was Tanx created?
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Does sin x tan x plus cos x equals tan x?
NO, sinxtanx=sinxsinx/cosx since tanx is sinx/cosx this is sin^2xcosx now add cosx cosx(sin^2x+1) after factoring Does this equal tanx? No, since this would require tanx to equal cosx(sin^2x+1) and it does not.
What is the Derivative of -x plus tanx?
(-x+tanx)'=-1+(1/cos2x)
