Mar 26, 2017 · Solve the trigonometric equation #tan^2x+tanx-1=0# in general and in the interval # [0,2pi)#?

Jan 4, 2016 · The trick is to remember that pi=180^o (-2pi)/3= (-2xx180^0)/3=-120^o hope that helped

(4pi)/3; (11pi)/6 4cos^2 x - 2 = sec^2 x - tan^2 x (1) Develop the right side: RS = 1/ (cos^2 x) - sin^2 x/ (cos^2 x) = (1 - sin^2 x)/ (cos^2 x) = cos^2 x/ (cos^2 x) = 1 The equation (1) becomes: 4cos^2 x - 2 = …

a. cos 2x = 0 Unit circle gives: 2x = pi/2 + 2kpi and 2x = (3pi)/2 + 2kpi 2x = pi/2 + 2kpi --> x = pi/4 + kpi 2x = (3pi)/2 + 2kpi --> x = (3pi)/4 + kpi Answers for (0, 2pi): pi/4; (3pi)/4; (5pi)/4; (7pi)/4 b. cos x = cos …

Solve both equations in terms of x: x=pi/9+ (2kpi)/3,kinZZand0≤ x< 2pi or x= (5pi)/9+ (2kpi)/3,kinZZand0≤ x< 2pi. Substitute k=0,1,2 to both of the equations to find x=pi/9, (5pi)/9, (7pi)/9, …

Let's choose (2pi)/3since we are in the second quadrant. Pay attention that -pi/3 is in the fourth quadrant, and this is wrong.

x=0,2pi Your question is cos (x-pi/6) + cos (x+pi/6) = sqrt3 in the interval [0,2pi].

Given r->"radius of the tyre"=0.3m" alpha->"constant angular acceleration of the tyre"=0.5(rad)/s^2 omega_0->"Initial angular velocity"=0 If theta represents the angular displacement when angular …

x = pi - sin^-1((-9+sqrt(57))/4)+ 2pin ;n in ZZ and x = 2pi + sin^-1((-9+sqrt(57))/4)+ 2pin ;n in ZZ Given: 2sin^2(x)+9sin(x)+3=0 Use the quadratic formula: sin(x ...

cos (180/7) cos (720/7) cos (900 /7 ) =cos (pi/7) cos ( (4pi)/7) cos ( (5pi) /7 ) =cos (pi/7) cos ( (5pi)/7) cos ( (4pi) /7 ) =cos (pi/7) cos (pi- (2pi)/7) cos ( (4pi ...