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How do you solve 1- 2cos squared theta without a calculator?
-1
How would you show all work for 1 minus 2cos squared theta?
You could not.1 - 2cos2theta is an expression which cannot be solved nor evaluated without information about theta. If you need to show its equivalence to another expression then it is necessary to know what the other expression is.
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 many solutions does cos squared minus two cos equal three?
One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3
What is the differenciation of sinx plus sin2x?
d/dx (sin x + sin 2x) = cos x + 2cos 2x
How do you solve 1- 2cos squared theta without a calculator?
-1
How would you show all work for 1 minus 2cos squared theta?
You could not.1 - 2cos2theta is an expression which cannot be solved nor evaluated without information about theta. If you need to show its equivalence to another expression then it is necessary to know what the other expression is.
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 many solutions does cos squared minus two cos equal three?
One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3
How do you Verify 1-sin to the 4th theta equals cos to the 4th theta plus 2cos to the 2nd theta times sin to the 2nd theta?
1 - sin4theta = cos4theta + 2cos2thetasin2theta. L: difference of squares. R: factor out cos2theta. (1 + sin2theta)(1 - sin2theta) = (cos2theta)(cos2theta + 2sin2theta)L: 1 -sin2theta always equals cos2theta. R: backwards from what I just said: cos2theta always equals 1 - sin2theta. (1 + sin2theta)(cos2theta) = (cos2theta)(1 - sin2theta + 2sin2theta)L: Nothing more to do. R: Just simplify what's inside the right parenthesees. (1 + sin2theta)(cos2theta) = (cos2theta)(1 + sin2theta) Done. L means left side of the equation; R means right side.
What is the name of the shape graphed by the function r = 2cos(3theta)?
Rose with 3 petals
How would you solve the following over the interval 0 is less than or equal to x which is less than 2pi here's the problem 2cos squared x minus 5cos x minus 3 equals 0?
2 cos2(x) - 5 cos(x) - 3 = 0To save some writing, let's temporarily write "Q" instead of "cos(x)".2 Q2 - 5 Q - 3 = 0Factor the trinomial:(2 Q + 1) (Q - 3) = 02Q + 1 = 02Q = -1Q = - 1/2Q - 3 = 0Q = 3Recalling that 'Q' is a cosine of an angle, (Q = -1/2) is the root that makes sense,because the cosine of a real angle can never be 3 .So now, we can get rid of 'Q' and go back to the cosine:cos(x) = -1/2x = 120 degreesx = 240 degrees
How do you differentiate cos squared x?
use the double angle formula for cos(2x) which is: cos(2x)=2cos^2(x)-1 by this relation cos^2(x)=(cos(2x)+1)/2 now we'd integrate this instead this will give sin(2x)/4+x/2 =) hope this helps
What is the differenciation of sinx plus sin2x?
d/dx (sin x + sin 2x) = cos x + 2cos 2x
What is the derivative of 2cosx times the inverse of sinx?
d/dx (2cos(x)sin⁻¹(x)) right, 2cos(x)sin⁻¹(x) is a product, so we'll differentiate using the product rule: d/dx(uv) = u d/dx(v) + v d/dx(u) u = 2cos(x) v = sin⁻¹(x) d/dx(u) = -2sin(x) to find d/dx(sin⁻¹(x)) we'll set y=sin⁻¹(x) x=sin(y) dx/dy = cos(y) dy/dx = 1/(cos(y)) cos²(y) + sin²(y)=1, dy/dx = 1/sqrt(1-sin²y) = 1/sqrt(1-x²) [x=sin(y)] so plugging all this into our product rule, d/dx (2cos(x)sin⁻¹(x)) = 2cos(x)/sqrt(1-x²) - 2sin(x)sin⁻¹(x).
What is the derivative of -2sin x?
The derivative of sin(x) is cos(x). Coefficients act like constants and always remain in derivatives. So, the derivative is -2cos(x).
Integrate sin x cos 2x?
By Angle-Addition, cos(2x) = 2cos(x)^2-1 So, sin(x)cos(2x) = [2cos(x)^2-1]sin(x) = 2sin(x)cos(x)^2 - sin(x) Int[2sin(x)cos(x)^2 - sin(x)] = (-2/3)cos(x)^3 + cos(x) +K
