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- Equation:
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Identical expressions
- 6cos^ two x+11cos^2+ four = zero
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Similar expressions
- 6cos^2x+11cos^2-4=0
- 6cos^2x-11cos^2+4=0
6cos^2x+11cos^2+4=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
2 2 6*cos (x) + 11*cos (x) + 4 = 0
$$\left(6 \cos^{2}{\left(x \right)} + 11 \cos^{2}{\left(x \right)}\right) + 4 = 0$$
Detail solution
Given the equation
$$\left(6 \cos^{2}{\left(x \right)} + 11 \cos^{2}{\left(x \right)}\right) + 4 = 0$$
transform
$$17 \cos^{2}{\left(x \right)} + 4 = 0$$
$$\left(6 \cos^{2}{\left(x \right)} + 11 \cos^{2}{\left(x \right)}\right) + 4 = 0$$
Do replacement
$$w = \cos{\left(x \right)}$$
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
$$a = 17$$
$$b = 0$$
$$c = 4$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$w_{1} = \frac{2 \sqrt{17} i}{17}$$
$$w_{2} = - \frac{2 \sqrt{17} i}{17}$$
do backward replacement
$$\cos{\left(x \right)} = w$$
Given the equation
$$\cos{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
substitute w:
$$x_{1} = \pi n + \operatorname{acos}{\left(w_{1} \right)}$$
$$x_{1} = \pi n + \operatorname{acos}{\left(\frac{2 \sqrt{17} i}{17} \right)}$$
$$x_{1} = \pi n + \frac{\pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
$$x_{2} = \pi n + \operatorname{acos}{\left(w_{2} \right)}$$
$$x_{2} = \pi n + \operatorname{acos}{\left(- \frac{2 \sqrt{17} i}{17} \right)}$$
$$x_{2} = \pi n + \frac{\pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
$$x_{3} = \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi$$
$$x_{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{2 \sqrt{17} i}{17} \right)}$$
$$x_{3} = \pi n - \frac{\pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
$$x_{4} = \pi n + \operatorname{acos}{\left(w_{2} \right)} - \pi$$
$$x_{4} = \pi n - \pi + \operatorname{acos}{\left(- \frac{2 \sqrt{17} i}{17} \right)}$$
$$x_{4} = \pi n - \frac{\pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
$$\left(6 \cos^{2}{\left(x \right)} + 11 \cos^{2}{\left(x \right)}\right) + 4 = 0$$
transform
$$17 \cos^{2}{\left(x \right)} + 4 = 0$$
$$\left(6 \cos^{2}{\left(x \right)} + 11 \cos^{2}{\left(x \right)}\right) + 4 = 0$$
Do replacement
$$w = \cos{\left(x \right)}$$
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
$$a = 17$$
$$b = 0$$
$$c = 4$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (17) * (4) = -272
Because D<0, then the equation
has no real roots,
but complex roots is exists.
w1 = (-b + sqrt(D)) / (2*a)
w2 = (-b - sqrt(D)) / (2*a)
or
$$w_{1} = \frac{2 \sqrt{17} i}{17}$$
$$w_{2} = - \frac{2 \sqrt{17} i}{17}$$
do backward replacement
$$\cos{\left(x \right)} = w$$
Given the equation
$$\cos{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
substitute w:
$$x_{1} = \pi n + \operatorname{acos}{\left(w_{1} \right)}$$
$$x_{1} = \pi n + \operatorname{acos}{\left(\frac{2 \sqrt{17} i}{17} \right)}$$
$$x_{1} = \pi n + \frac{\pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
$$x_{2} = \pi n + \operatorname{acos}{\left(w_{2} \right)}$$
$$x_{2} = \pi n + \operatorname{acos}{\left(- \frac{2 \sqrt{17} i}{17} \right)}$$
$$x_{2} = \pi n + \frac{\pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
$$x_{3} = \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi$$
$$x_{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{2 \sqrt{17} i}{17} \right)}$$
$$x_{3} = \pi n - \frac{\pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
$$x_{4} = \pi n + \operatorname{acos}{\left(w_{2} \right)} - \pi$$
$$x_{4} = \pi n - \pi + \operatorname{acos}{\left(- \frac{2 \sqrt{17} i}{17} \right)}$$
$$x_{4} = \pi n - \frac{\pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
Rapid solution
[src]
/ ____\
pi |2*\/ 17 |
x1 = -- - I*asinh|--------|
2 \ 17 /
$$x_{1} = \frac{\pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
/ ____\
pi |2*\/ 17 |
x2 = -- + I*asinh|--------|
2 \ 17 /
$$x_{2} = \frac{\pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
/ ____\
3*pi |2*\/ 17 |
x3 = ---- - I*asinh|--------|
2 \ 17 /
$$x_{3} = \frac{3 \pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
/ ____\
3*pi |2*\/ 17 |
x4 = ---- + I*asinh|--------|
2 \ 17 /
$$x_{4} = \frac{3 \pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}$$
x4 = 3*pi/2 + i*asinh(2*sqrt(17)/17)
Sum and product of roots
[src]
sum
/ ____\ / ____\ / ____\ / ____\ pi |2*\/ 17 | pi |2*\/ 17 | 3*pi |2*\/ 17 | 3*pi |2*\/ 17 | -- - I*asinh|--------| + -- + I*asinh|--------| + ---- - I*asinh|--------| + ---- + I*asinh|--------| 2 \ 17 / 2 \ 17 / 2 \ 17 / 2 \ 17 /
$$\left(\left(\frac{3 \pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}\right) + \left(\left(\frac{\pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}\right) + \left(\frac{\pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}\right)\right)\right) + \left(\frac{3 \pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}\right)$$
=
4*pi
$$4 \pi$$
product
/ / ____\\ / / ____\\ / / ____\\ / / ____\\ |pi |2*\/ 17 || |pi |2*\/ 17 || |3*pi |2*\/ 17 || |3*pi |2*\/ 17 || |-- - I*asinh|--------||*|-- + I*asinh|--------||*|---- - I*asinh|--------||*|---- + I*asinh|--------|| \2 \ 17 // \2 \ 17 // \ 2 \ 17 // \ 2 \ 17 //
$$\left(\frac{\pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}\right) \left(\frac{\pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}\right) \left(\frac{3 \pi}{2} - i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}\right) \left(\frac{3 \pi}{2} + i \operatorname{asinh}{\left(\frac{2 \sqrt{17}}{17} \right)}\right)$$
=
/ ____\
2 2|2*\/ 17 |
/ ____\ 4 5*pi *asinh |--------|
4|2*\/ 17 | 9*pi \ 17 /
asinh |--------| + ----- + ----------------------
\ 17 / 16 2
$$\operatorname{asinh}^{4}{\left(\frac{2 \sqrt{17}}{17} \right)} + \frac{5 \pi^{2} \operatorname{asinh}^{2}{\left(\frac{2 \sqrt{17}}{17} \right)}}{2} + \frac{9 \pi^{4}}{16}$$
asinh(2*sqrt(17)/17)^4 + 9*pi^4/16 + 5*pi^2*asinh(2*sqrt(17)/17)^2/2
Numerical answer
[src]
x1 = 1.5707963267949 - 0.4678194397177*i
x2 = 1.5707963267949 + 0.4678194397177*i
x3 = 4.71238898038469 - 0.4678194397177*i
x4 = 4.71238898038469 + 0.4678194397177*i
x4 = 4.71238898038469 + 0.4678194397177*i