Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2-8*x+15=0
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Equation -x^2+4x+3=x^2-x-(1+2x^2)
-
Equation 3x+3=5x
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Equation x^2+3x-4=0
- Express {x} in terms of y where:
- -17*x+2*y=9
- -13*x+7*y=-15
- -14*x-8*y=20
- -1*x+4*y=-10
- Derivative of:
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2sin4x
- Integral of d{x}:
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2sin4x
- Graphing y =:
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2sin4x
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Identical expressions
- two sin4x=6sin^2*2x- one
- 2 sinus of 4x equally 6 sinus of squared multiply by 2x minus 1
- two sinus of 4x equally 6 sinus of squared multiply by 2x minus one
- 2sin4x=6sin2*2x-1
- 2sin4x=6sin²*2x-1
- 2sin4x=6sin to the power of 2*2x-1
- 2sin4x=6sin^22x-1
- 2sin4x=6sin22x-1
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Similar expressions
- 2sin4x=6sin^2*2x+1
- y=e^5*x*(c1*cos(4*x)+c2*sin(4*x))
- 2sin^4(x)+3cos(2x)+1=0
2sin4x=6sin^2*2x-1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2*sin(4*x) = 6*sin (2)*x - 1
$$2 \sin{\left(4 x \right)} = x 6 \sin^{2}{\left(2 \right)} - 1$$
Detail solution
Given the equation
$$2 \sin{\left(4 x \right)} = x 6 \sin^{2}{\left(2 \right)} - 1$$
transform
$$- 6 x \sin^{2}{\left(2 \right)} + 2 \sin{\left(4 x \right)} = 0$$
$$- x 6 \sin^{2}{\left(2 \right)} + 2 \sin{\left(4 x \right)} = 0$$
Do replacement
$$w = \sin{\left(2 \right)}$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = - 6 x$$
$$b = 0$$
$$c = 2 \sin{\left(4 x \right)}$$
, then
The equation has two roots.
or
$$w_{1} = - \frac{\sqrt{3} \sqrt{x \sin{\left(4 x \right)}}}{3 x}$$
$$w_{2} = \frac{\sqrt{3} \sqrt{x \sin{\left(4 x \right)}}}{3 x}$$
do backward replacement
$$\sin{\left(2 \right)} = w$$
substitute w:
$$2 \sin{\left(4 x \right)} = x 6 \sin^{2}{\left(2 \right)} - 1$$
transform
$$- 6 x \sin^{2}{\left(2 \right)} + 2 \sin{\left(4 x \right)} = 0$$
$$- x 6 \sin^{2}{\left(2 \right)} + 2 \sin{\left(4 x \right)} = 0$$
Do replacement
$$w = \sin{\left(2 \right)}$$
This equation is of the form
a*w^2 + b*w + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = - 6 x$$
$$b = 0$$
$$c = 2 \sin{\left(4 x \right)}$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (-6*x) * (2*sin(4*x)) = 48*x*sin(4*x)
The equation has two roots.
w1 = (-b + sqrt(D)) / (2*a)
w2 = (-b - sqrt(D)) / (2*a)
or
$$w_{1} = - \frac{\sqrt{3} \sqrt{x \sin{\left(4 x \right)}}}{3 x}$$
$$w_{2} = \frac{\sqrt{3} \sqrt{x \sin{\left(4 x \right)}}}{3 x}$$
do backward replacement
$$\sin{\left(2 \right)} = w$$
substitute w: