2018年5月2日 · I need help solving this problem, Solve the following trigonometric equation on the interval [ 0, 2pi ) : 2cos2x (sin2x) + sin2x = 0?

Use Trig properties and/or algebra to transform the given equation into one or more equations of the form: TrigFunction (Something) = SomeNumber Find the general solution (s) to each …

Question 1137706: solve 2cos2x-2cosx=0 over the interval [0,2pi) Answer by MathLover1 (20820) (Show Source):

2018年3月25日 · Use The Pythagorean Identity sin 2 θ+cos 2 θ=1, so sin 2 θ=1-cos 2 θ. Also sin 2 45 = cos 2 45 = 1/2

Joy H. asked • 01/25/16 How do you solve 2cos^2x-1=0?How do you solve 2cos^2x-1=0?

First use the double angle formula to get 2cos x+2 (2cos^2 x-1)=0 then distribute and simplify by dividing by 2 to get 2cos^2 x+cos x-1=0 then factor into (2cos x-1) (cos x+1)=0 and solve for x …

Question 1176499: Find the value of x in the range 0°≤ x ≤ 360°, which satisfy the equation below. (i) sin3xsinx = 2cos2x + 1 (ii) 1 + cosx = 2sin²x (iii) 3cotx + tanx - 4 = 0 Answer by Boreal …

2018年4月13日 · Select your answer from the choices below: A) 45°, 315° B) 45°, 135° C) 45°, 135°, 225°, 315° D) 45°, 225°Solve 2cos2x - 1 = 0 for 0° < x < 360°

The period is the time frame it takes the function to complete one cycle. In other words, when does the graph revisit its starting peak? You can calculate the period by dividing 2π by the …

Jo F. asked • 02/06/14 please, prove that 1-2cos^2x = tan^2x-1/tan^2x+1