Find all the solutions in the interval [0, 2pi): 2 sin 3x cos 3x = 1?
2sin3xcos3x=1 sin6x=1 6x= 2n(pi)+ pi/2 x=(4n+1)pi/12 put different values of n keeping in mind that 0
Find all solutions in the interval [0,2pi): 3 tan x-3=0?
3tan x - 3 = 0 3tan x = 3 tan x = 3/3 tan x = 1 x = tanˉ¹ (1) x = π/4, 5π/4 radians (45°, 225°)
How can I find all the solutions for sin (3x - pi/4) =1 that are the intervals of [0,2pi]?
sin(x) = 1 over [0,2π] when x = π/2. So the challenge is to find when 3x - π/4 = π/2. So solving, 3x = 3π/4, x = π/4.To wrap up, sin(3(π/4) - π/4) = sin(π/2) = 1.
Find all the solutions in the interval [0, 2pi): 5 sin^2 x + 8 sin x - 4= 0?
here's a neat trick to use let x=sinx then you have 5x^2+8x-4=0 (5x-2)(x+2) then get x by itself x=2/5 x=-2 so when does sin x=2/5 sin x=-2?
Find all the solutions in the interval [0, 2pi): 2 sin^2 (x/4) - 3 cos (x/4) = 0?
Using the identity sin^2 + cos^2 = 1, we can turn the transcendental equation given into a quadratic in terms of cos^2(x/4): Since sin^2 (x/4) = 1 - cos^2 (x/4), we can substitute 1-cos^2(x/4) for sin2(x/4) in the given equation to give us 2 (1 - cos^2 (x/4) ) - 3 cos (x/4) = 0 or -2 cos^2 (x/4) - 3 cos(x/4) + 2 = 0 (-2 cos(z/4) + 1) (cos(z/4) + 2) = 0 As cos(z/4) is always greater than or equal to -1 for any real valued z, the second factor can't equal zero, so all roots must satisfy -2 cos(z/4) + 1 = 0 1 = 2 cos (z/4) cos (z/4) = 1/2 If z is on the interval [0, 2pi), then z/4 is on the interval [0, pi/2), limiting us to just one root: z/4 = pi/3 so z = 4 pi / 3 Your answer is (b)
Find all solutions for sin2x-tanx=0 in interval [0,2pi) ?
If sin2x - tanx = 0, then adding tanx to both sides means that sin2x = tanx, or that 2sinxcosx = sinx/cosx. Dividing both sides by sinx, we have 2cosx = 1/cosx. Then multiplying both sides by cosx we have 2cos^x = 1, or cosx = plus or minus the square root of 1/2. You can solve it pretty easily for X at that point.
Find all solutions in the interval [ 0 , 2 ). 18 tan3 x = 6 tan x?
if you mean 18 (tan x)^3 = 6 tan x then 18 (tan x)^3 - 6 tan x = 0 6 tan x (3(tan x)^2 - 1) = 0 tan x = 0, x = 0 3(tan x)^2 - 1 = 0, (tan x)^2 = 1/3 tan x = sqrt(3)/3, x = pi/6 so the solution is x = 0, pi/6 in the interval [ 0 , 2 ).
How do I find all the solutions in the interval [0, 2pi] of sin^2x+3cos^2=0?
(sinx)^2 + 3(cosx)^2 = 0(sinx)^2 +3[1-(sinx)^2] = 03 - 2(sinx)^2 = 0sinx = (3^0.5)/2 or -(3^0.5)/2therefore x = 60, 120, 240, 300x = pi/3, (2pi)/3, (4pi)/3, (5pi)/3
Find all the solutions in the interval [0, 2pi) : csc x + 2 = 0.?
csc(x) + 2 = 0 1/sin(x) + 2 2sin(x) = -1 sin(x) = -1/2 The sin curve is negative from the interval of [pi, 2pi] and sin gives a value of 1/2 only when the reference angle is pi/6 So your two answers are 7pi/6 and 11pi/6 - d
Find all solutions of the equation interval [0,2 pi)?
1) 2s^2 +3s +1=0 ie (2s+1)(s+1)=0 ie s=-1/2 or -1 so sinx= -1/2 or -1 sinx= -1/2 gives x=7pi/6, 11pi/6 2)using the identity, cos^2 x + sin^2 x = 1 wecan rewwrite the equation, 2(1-cos^2 x) = 3+cosx this simplifies to, 2cos^2 x + cosx +1 =0 2c^2 + c +1 = 0 and this equation has no solution as the discriminant is negative.