The solution set of the equation tan3x = TaNx is the thinking and process of solving

TaNx increases in a cycle
So we have 3x = x + K π
x=kπ/2
And Tan π / 2 does not exist
So K is even
So x = m π, m ∈ Z

The limit of TaNx / tan3x, X tends to π / 6 Sorry, it's x approaching 0

By substituting equivalent infinitesimal, limtanx / tan3x = limx / 3x = 1 / 3 (x tends to 0)

The solution set of the equation tan3x = TaNx is Why should not 3x = x + K π be 3x = x + K π

3X=X+2kπ
x=kπ
For 3x = x + (2k + 1) π
X = (2k + 1) π / 2 (round off)
Because y = TaNx, the domain x ≠ π / 2 + K π

Solving the limit of (TaNx) / (tan3x) when x → Pai / 2

The law of lobida can be used
Derivation up and down
=(secx)^2/3(sec3x)^2
=(1/3)*(cos3x)^2/(cosx)^2
This is the 0 / 0 type. You can use the law of l'urbida
lim(x→π/2)(cos3x)^2/(cosx)^2
=[LIM (x → π / 2) (cos3x) / (cosx)] ^ 2 (here is the square of the entire limit)
=[lim(x→π/2)-3sin3x/(-sinx)]^2
lim(x→π/2)-3sin3x=-3
lim(x→π/2)-sinx=-1
So [LIM (x → π / 2) - 3sin3x / (- SiNx)] ^ 2
=3^2=9
So the original limit = (1 / 3) * 9 = 3

The proof of cosinx + cosinx How to prove this

Molecule = sin (3x + x) + sin (3x-x)
=sin3xcosx+cos3xsinx+sin3xcosx-cos3xsinx
=2sin3xcosx
Similarly, denominator = 2cos3xcosx
So left = sin3x / cos3x = tan3x = right

Simplify (1 + sin4x + cos4x) / (1 + sin4x-cos4x)

0

f(sinx)=cos2x=1-2(sinx)^2
f(x)=1-2x^2
f(cosx)=1-2(cosx)^2=-cos2x

Indefinite integral of the square of 1 + cosx / 1 + SiNx The denominator is the square of SiNx plus 1

The original formula = ∫ (1 / 2-cos ^ 2) DX + ∫ (1 / (1 + sin ^ 2x)) d (SiNx) = ∫ [1 / (COS ^ 2 (2sec ^ 2x-1))] DX + arctan (SiNx) = ∫ (1 / (1 + 2tan ^ 2x)) d (TaNx) + arctan (SiNx) = [(√ 2) / 2] arctan [(√ 2) (TaNx)] + arctan (SiNx) + C

Why sin ^ 2x = 1 / 2 (1-cos2x)

Double angle formula
cos2x=1-2sin²x
So sin? X = 1 / 2 (1-cos2x)

Y = - cos2x = sin (2x + 3 π / 2) why?

sin(2x+3π/2)=sin2xcos3π/2+cos2xsin3π/2
cos3π/2=0,sin3π/2=-1
Original formula = - cos2x

The solution set of the equation tan3x = TaNx is the thinking and process of solving