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Hello-sirs-i-want-that-you-explain-me-how-we-solve-the-trigonometric-inequality-with-tangente-we-can-take-for-example-tan2x-3-i-can-solve-this-type-of-equality-but-not-the-inequality-Pleas




Question Number 77472 by mathocean1 last updated on 06/Jan/20
Hello sirs...   i want that you explain me how    we solve the trigonometric  inequality with tangente.   we can take for example tan2x≥(√3)  i can solve this type of equality  but not the inequality! Please   i need your help... Even the steps  to make graphic to read the solu−  tion.
Hellosirsiwantthatyouexplainmehowwesolvethetrigonometricinequalitywithtangente.wecantakeforexampletan2x3icansolvethistypeofequalitybutnottheinequality!PleaseineedyourhelpEventhestepstomakegraphictoreadthesolution.
Commented by MJS last updated on 06/Jan/20
(1) solve the equality for x within one period       tan 2x =(√3); −(π/4)≤x<(π/4)       ⇒ x=(π/6)  (2) calculate the gradient of the curve       (d/dx)[tan 2x]=(2/(cos^2  2x))>0∀x∈R\{−(π/4)+(π/2)n}; n∈Z       also tan 2x is not defined for x=−(π/4)+(π/2)n       ⇒ within this period tan 2x ≥(√3) with (π/6)≤x<(π/4)       ⇒ tan 2x ≥(√3) with (π/6)+(π/2)n≤x<(π/4)+(π/2)n; n∈Z
(1)solvetheequalityforxwithinoneperiodtan2x=3;π4x<π4x=π6(2)calculatethegradientofthecurveddx[tan2x]=2cos22x>0xR{π4+π2n};nZalsotan2xisnotdefinedforx=π4+π2nwithinthisperiodtan2x3withπ6x<π4tan2x3withπ6+π2nx<π4+π2n;nZ
Commented by mathocean1 last updated on 06/Jan/20
Thank very much...
Thankverymuch
Commented by mathocean1 last updated on 06/Jan/20
I haven”t studied about gradient  of curve sir... what is it?
Ihaventstudiedaboutgradientofcurvesirwhatisit?
Commented by MJS last updated on 06/Jan/20
the gradient in a point of the curve is the  value of the derivate  P= ((p),((f(p))) ) ⇒ gradient = f ′(p)  f ′(p)>0 ⇒ curve is increasing in P  f ′(p)=0 ⇒ curve is flat in P  f ′(p)<0 ⇒ curve is decreasing in P
thegradientinapointofthecurveisthevalueofthederivateP=(pf(p))gradient=f(p)f(p)>0curveisincreasinginPf(p)=0curveisflatinPf(p)<0curveisdecreasinginP
Commented by mathocean1 last updated on 06/Jan/20
Thank you...
Thankyou

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