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How do you solve 1 minus 2cos squared theta?
Since there is no equation, there is nothing that can be solved.
How do you solve 1- 2cos squared theta without a calculator?
-1
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 minus 2cos squared theta?
Since there is no equation, there is nothing that can be solved.
How do you solve 1- 2cos squared theta without a calculator?
-1
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 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
Find a model for simple harmonic motion if the position at t equals 0 is 2 the amplitude is 2 centimeters and the period is 6 seconds?
I'm not entirely sure about this, but imagine that you have a coordinate system where the x-axis is in units of seconds (variable name t) and the y-axis is in units of centimeters (variable name d (for displacement)). Simple harmonic motion would indicate a simple periodic motion pattern, which would hint at a sine or cosine function. Using the above coordinate system, an unmodified cosine wave would have an amplitude of 1.0 cm and a period of 2(pi) seconds: d=cos(t) I chose cosine because cosine at t=0 (x=0) would be at whatever its amplitude is. To get the amplitude to be 2.0 cm is easy, you must simply modify the amplitude of the cosine function by multiplying it by 2. So far: d=2cos(t) To modify the period to fit the problem statement, you must start with the fact that an unmodified cosine function would have a period of 2(pi). If you were to cut the period in half, you would multiply the argument inside the function by 2, which would give it a period of (pi). This would look like: y=cos(2x) So, if multiplying by 2 changes the period by 1/2, what would we need to multiply by to get a period of 6 seconds? 6 seconds is 6/2(pi)=3/(pi) of 2(pi), to get 3/(pi) of the current period, we must multiply the interior argument by (pi)/3. This gives us a final model of: d=2cos((pi/3)t) If this model is correct, one period of the cosine wave will complete by t=6 seconds. This means that the model will return values of 2 at t=0 and t=6. Let's check: d(0)=2cos((pi/3)(0))=2cos(0)=2*1=2 d(6)=2cos((pi/3)(6))=2cos(2(pi))=2*1=2 These values check out. This means that our model works. So, the final answer is: d=2cos((pi/3)t)
What is the differenciation of sinx plus sin2x?
d/dx (sin x + sin 2x) = cos x + 2cos 2x
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
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).
