Jun 11, 2018 · 2cosx −1 = 0 ⇒ cosx = 1 2. since cosx > 0 x in first/fourth quadrant. x = cos−1( 1 2) = 60∘ ← first quadrant. or x = (360 − 60)∘ = 300∘ ← fourth quadrant. x ∈ {60,180,300} → 0 ≤ x < 360. Answer link. See below Given 1+cosx-2sin^2x=0 we can do some changes based on trigonometric identities like 1+cosx-2 (1-cos^2x)=0 1
The anti-derivative of cos 2x is nothing but the integral of cos 2x. We know that the integral of cos x is sin x + C. Using the formula of integration ∫cos (ax + b) = (1/a) sin (ax + b) + C, the anti-derivative of cos 2x is (1/2) sin 2x + C, where C is constant of integration. Hence, we have obtained the anti-derivative of cos 2x as (1/2) sin
Apr 23, 2015 · The formula can be proven by applying: 1) Least common multiple; 2) applying the trigonometric entity sin^2x + cos^2x=1 Head Key-relation : sin^2x + cos^2x=1 Key-concept: Least common multiple; when no common multiples, just multiply the terms in the denominator. Calculation The above formula can be proven by transforming left side to right side: 1/(1-sin x)+1/(1+sin x)= (1+sin x + 1-sin x
If tan(α+iβ) =eiθ ; where α,β ∈ R, θ ≠(2n+1)π 2,n ∈ Z and i =√−1, then. If the equation 4sin(x+ π 3)cos(x − π 6) =a2 +√3sin2x−cos2x has a solution, then the value of a can be. If f (x) = cosx [x π]+ 1 2, where x is not an integral multiple of π and [.] denotes the greatest integer function, then. Jun 11, 2018 · Consider the integral I = ∫ xsinx \1 + cos^2x dx, x∈[0,π] (i) Express I = π/2 ∫ sinx/1 + cos^2x dx, x∈[0,π] (ii) Show that I = π^2/4 asked Jan 18, 2021 in Integrals by Sadhri ( 29.7k points) May 3, 2015 · Another way: #cos 2x = 2.cos^2 x - 1 = 1# #cos^2 x = 1# cos x = 1 -> x = 0 and x = 2pi cos x = -1 -> x = pi. Check: x = pi -> 2x = 2pi -> cos 2x = 1 -> 1 = 1 Correct. Sdpam.