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How do you solve 6cos2 theta plus 5cos theta minus 4 equal to 0?
Assume the first term is cos-squared(theta) rather than cos(2*theta). 6cos2(t) + 5cos(t) - 4 = 0 [6cos2(t) + 8cos(t) - 3 cos(t) - 4] = 0 [3cos(t) + 4]*[2cos(t) - 1] = 0 which gives 3cos(t) = -4 so that cos(t) = -4/3 or 2cos(t) = 1 so that cos(t) = 1/2 The first of these is clearly not a possible solution, whereas the second can hve one or more solutions, depending on the domain - which is not specified in the question. If the initial assumption is incorrect and the first term WAS 6cos(2*theta) then using the formula for double angles: cos(2(t) = cos2(t) - sin2(t) = cos2(t) - [1 - cos2(t)] = 2cos2(t) - 1 and so, the equation becomes 6*[2cos2(t) - 1] + 5cos(t) - 4 = 0 or 12cos2(t) + 5cos(t) - 10 = 0 Solve this quadratic equation for cos(t) and then find the values of t (or theta) within the domain specified.
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
Is 1- cos 2 x 1 plus cos 2 x equals sin squared x cos squared x an identity?
No, (sinx)^2 + (cosx)^2=1 is though
How do you solve cos squared x - cosx equals 2 for 0x2pi?
cos2(x) - cos(x) = 2 Let y = cos(x) then y2 - y = 2 or y2 - y - 2 = 0 factorising, (y - 2)(y + 1) = 0 that is y = 2 or y = -1 Substitutng back, this would require cos(x) = 2 or cos(x) = -1 But cos(x) cannot be 2 so cos(x) = -1 Then x = cos-1(-1) => x = pi radians.
How do you solve 6cos2 theta plus 5cos theta minus 4 equal to 0?
Assume the first term is cos-squared(theta) rather than cos(2*theta). 6cos2(t) + 5cos(t) - 4 = 0 [6cos2(t) + 8cos(t) - 3 cos(t) - 4] = 0 [3cos(t) + 4]*[2cos(t) - 1] = 0 which gives 3cos(t) = -4 so that cos(t) = -4/3 or 2cos(t) = 1 so that cos(t) = 1/2 The first of these is clearly not a possible solution, whereas the second can hve one or more solutions, depending on the domain - which is not specified in the question. If the initial assumption is incorrect and the first term WAS 6cos(2*theta) then using the formula for double angles: cos(2(t) = cos2(t) - sin2(t) = cos2(t) - [1 - cos2(t)] = 2cos2(t) - 1 and so, the equation becomes 6*[2cos2(t) - 1] + 5cos(t) - 4 = 0 or 12cos2(t) + 5cos(t) - 10 = 0 Solve this quadratic equation for cos(t) and then find the values of t (or theta) within the domain specified.
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
2 cos x plus 1 cos x plus -1 equals 0?
To solve the equation 2cos(x) + cos(x) - 1 = 0, we first combine like terms to get 3cos(x) - 1 = 0. Then, we isolate the cosine term by adding 1 to both sides to get 3cos(x) = 1. Finally, we divide by 3 to solve for cos(x), which gives cos(x) = 1/3. Therefore, x = arccos(1/3) or approximately 70.53 degrees.
What is the solution to cos x equals 3 cos x - 2?
1
Is 1- cos 2 x 1 plus cos 2 x equals sin squared x cos squared x an identity?
No, (sinx)^2 + (cosx)^2=1 is though
What does negative sine squared plus cosine squared equal?
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
How do you solve cos squared x - cosx equals 2 for 0x2pi?
cos2(x) - cos(x) = 2 Let y = cos(x) then y2 - y = 2 or y2 - y - 2 = 0 factorising, (y - 2)(y + 1) = 0 that is y = 2 or y = -1 Substitutng back, this would require cos(x) = 2 or cos(x) = -1 But cos(x) cannot be 2 so cos(x) = -1 Then x = cos-1(-1) => x = pi radians.
What is the limit as x approaches 0 of 1 plus cos squared x over cos x?
2
Solve this equation. 2 plus 2 plus 2 plus 2?
The correct ansewer is 8
What is the derivative of 6 sin x plus 2 cos x?
F(x) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
