2014年11月6日 · Free Online Scientific Notation Calculator. Solve advanced problems in Physics, Mathematics and Engineering. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History.

2016年11月19日 · Use the identity cos^2beta = 1- 2sin^2beta. 1 - 2sin^2theta + costheta = 0 1 - 2 (1 - cos^2theta) + costheta = 0 1 - 2 + 2cos^2theta + costheta = 0 2cos^2theta + costheta - 1 = 0 2cos^2theta + 2costheta - costheta - 1 = 0 2costheta (costheta + 1) - (costheta + 1) = 0 (2costheta - 1) (costheta + 1) = 0 theta = pi/3 + 2pin, (2pi)/3 + 2pin, pi + 2pin. Hopefully this helps!

2015年4月14日 · Replace cos2x = 1 − 2sin2x: f (x) = cos2x + sinx = 1 − 2sin2x + sinx = 0 Call sinx = t. This is a quadratic equation in t: f (t) = − 2t2 +t + 1 = 0 Solve this quadratic equation. Because a + b + c = 0, one real root is t1 = 1 and the other is t2 = − 1 2 Next, solve the basic trig equation: t1 = sinx = 1 → x = π 2 Solve: t2 = sinx = − 1 2 → x = 7π 6 and x = 11π 6 Answers within ...

How do you find the exact values of sin 2u, cos 2u, and tan 2u using the double-angle formulas given tanu = 3 4, 0 <u <π 2?

2017年3月14日 · Here's an alternate answer. Recall the identity sin2θ +cos2θ = 1. If you rearrange for cos2θ, you should get cos2θ = 1 − sin2θ. Substituting, we have: sin2θ− (1 −sin2θ) = 0 sin2θ− 1 + sin2θ = 0 2sin2θ = 1 sin2θ = 1 2 sinθ = ± 1 √2 Now consider the 1 − 1 − √2 right triangle. This means that θ = π 4 +2πn, 3π 4 +2πn, 5π 4 +2πn and 7π 4 + 2πn Note the …

2016年5月28日 · cos2x − cosx = 0 cosx(1 − cosx) = 0 cosx = 0;cosx = 1 x = 0o; x = 90o Test: (a) x = 0o cos20 - cos0 12 −1 = 0 (b) x = 0o cos290 - cos90 02 −0 = 0

2016年10月21日 · This is a trivial quadratic equation in cosx 4cos2x − 1 = 0 ∴ 4cos2x = 1 ∴ cos2x = 1 4 ∴ cosx = ± 1 2 In order to solve this you need to be familiar with the graphs of the trig functions along with common values: The range of soutions is not specified in the question, so lets assume we want a solutio in radians, and −2π ≤ x ≤ 2π

2016年6月2日 · As you probably know, trinomials of the form y = ax^2 + bx + c, a != 1 can be solved by finding two numbers that multiply to a xx c and that add to b. :. Two numbers that multiply to -2 and that add to -1 are -2 and +1. (2cos^2theta - 2costheta) + (costheta - 1) = 0 Factor out a common factor from both pairs of binomials (encased in the ...

2016年4月18日 · Use the Property: cos 2A=2cos^-2A-1 cosx-(2cos^2x -1)=0 -1 [cosx -2cos^2x+1]=0 2cos^2x-cosx-1=0 (2cosx+1)(cosx-1)=0 cosx=-1/2 or cos x=1 x=cos^-1(-1/2) or x=cos^-1 1 ...

2015年9月15日 · theta=pi/3 vv theta= (5pi)/3 vv theta=pi cos2theta=cos^2theta-sin^2theta cos^2theta-sin^2theta+costheta=0 sin^2theta+cos^2theta=1 => sin^2theta=1-cos^2theta cos^2theta- (1-cos^2theta)+costheta=0 cos^2theta-1+cos^2theta+costheta=0 2cos^2theta+costheta-1=0 2cos^2theta-costheta+2costheta-1=0 costheta (2costheta …