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Equation (x-1)(x+9)=8x
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- Equation 3(tg^2(x)-1)*sqrt(-5cosx)=0
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Identical expressions
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Similar expressions
- 3(tg^2(x)+1)*sqrt(-5cosx)=0
- 3(tg^2(x)-1)*sqrt(5cosx)=0
3(tg^2(x)-1)*sqrt(-5cosx)=0 equation
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[src]
/ 2 \ ___________ 3*\tan (x) - 1/*\/ -5*cos(x) = 0
$$\sqrt{- 5 \cos{\left(x \right)}} 3 \left(\tan^{2}{\left(x \right)} - 1\right) = 0$$
Detail solution
Given the equation
$$\sqrt{- 5 \cos{\left(x \right)}} 3 \left(\tan^{2}{\left(x \right)} - 1\right) = 0$$
transform
$$3 \sqrt{5} \sqrt{- \cos{\left(x \right)}} \left(\tan^{2}{\left(x \right)} - 1\right) = 0$$
$$\sqrt{- 5 \cos{\left(x \right)}} 3 \left(\tan^{2}{\left(x \right)} - 1\right) = 0$$
Do replacement
$$w = \tan{\left(x \right)}$$
Expand the expression in the equation
$$\sqrt{5} \sqrt{- \cos{\left(x \right)}} \left(3 w^{2} - 3\right) = 0$$
We get the quadratic equation
$$3 \sqrt{5} w^{2} \sqrt{- \cos{\left(x \right)}} - 3 \sqrt{5} \sqrt{- \cos{\left(x \right)}} = 0$$
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
$$a = 3 \sqrt{5} \sqrt{- \cos{\left(x \right)}}$$
$$b = 0$$
$$c = - 3 \sqrt{5} \sqrt{- \cos{\left(x \right)}}$$
, then
The equation has two roots.
or
$$w_{1} = 1$$
$$w_{2} = -1$$
do backward replacement
$$\tan{\left(x \right)} = w$$
Given the equation
$$\tan{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{atan}{\left(w \right)}$$
Or
$$x = \pi n + \operatorname{atan}{\left(w \right)}$$
, where n - is a integer
substitute w:
$$x_{1} = \pi n + \operatorname{atan}{\left(w_{1} \right)}$$
$$x_{1} = \pi n + \operatorname{atan}{\left(1 \right)}$$
$$x_{1} = \pi n + \frac{\pi}{4}$$
$$x_{2} = \pi n + \operatorname{atan}{\left(w_{2} \right)}$$
$$x_{2} = \pi n + \operatorname{atan}{\left(-1 \right)}$$
$$x_{2} = \pi n - \frac{\pi}{4}$$
$$\sqrt{- 5 \cos{\left(x \right)}} 3 \left(\tan^{2}{\left(x \right)} - 1\right) = 0$$
transform
$$3 \sqrt{5} \sqrt{- \cos{\left(x \right)}} \left(\tan^{2}{\left(x \right)} - 1\right) = 0$$
$$\sqrt{- 5 \cos{\left(x \right)}} 3 \left(\tan^{2}{\left(x \right)} - 1\right) = 0$$
Do replacement
$$w = \tan{\left(x \right)}$$
Expand the expression in the equation
$$\sqrt{5} \sqrt{- \cos{\left(x \right)}} \left(3 w^{2} - 3\right) = 0$$
We get the quadratic equation
$$3 \sqrt{5} w^{2} \sqrt{- \cos{\left(x \right)}} - 3 \sqrt{5} \sqrt{- \cos{\left(x \right)}} = 0$$
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
$$a = 3 \sqrt{5} \sqrt{- \cos{\left(x \right)}}$$
$$b = 0$$
$$c = - 3 \sqrt{5} \sqrt{- \cos{\left(x \right)}}$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (3*sqrt(5)*sqrt(-cos(x))) * (-3*sqrt(5)*sqrt(-cos(x))) = 180*(-cos(x))
The equation has two roots.
w1 = (-b + sqrt(D)) / (2*a)
w2 = (-b - sqrt(D)) / (2*a)
or
$$w_{1} = 1$$
$$w_{2} = -1$$
do backward replacement
$$\tan{\left(x \right)} = w$$
Given the equation
$$\tan{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{atan}{\left(w \right)}$$
Or
$$x = \pi n + \operatorname{atan}{\left(w \right)}$$
, where n - is a integer
substitute w:
$$x_{1} = \pi n + \operatorname{atan}{\left(w_{1} \right)}$$
$$x_{1} = \pi n + \operatorname{atan}{\left(1 \right)}$$
$$x_{1} = \pi n + \frac{\pi}{4}$$
$$x_{2} = \pi n + \operatorname{atan}{\left(w_{2} \right)}$$
$$x_{2} = \pi n + \operatorname{atan}{\left(-1 \right)}$$
$$x_{2} = \pi n - \frac{\pi}{4}$$
Sum and product of roots
[src]
sum
pi pi - -- + -- 4 4
$$- \frac{\pi}{4} + \frac{\pi}{4}$$
=
0
$$0$$
product
-pi pi ----*-- 4 4
$$- \frac{\pi}{4} \frac{\pi}{4}$$
=
2 -pi ----- 16
$$- \frac{\pi^{2}}{16}$$
-pi^2/16
Rapid solution
[src]
-pi
x1 = ----
4
$$x_{1} = - \frac{\pi}{4}$$
pi
x2 = --
4
$$x_{2} = \frac{\pi}{4}$$
x2 = pi/4
Numerical answer
[src]
x1 = -79.3252145031423
x2 = -18.0641577581413
x3 = -85.6083998103219
x4 = -5.49778714378214
x5 = -55.7632696012188
x6 = 32.2013246992954
x7 = 55.7632696012188
x8 = 84.037603483527
x9 = -32.2013246992954
x10 = 24.3473430653209
x11 = 52.621676947629
x12 = 46.3384916404494
x13 = 14.9225651045515
x14 = 58.9048622548086
x15 = -25.9181393921158
x16 = -76.1836218495525
x17 = -19.6349540849362
x18 = 68.329640215578
x19 = 63.6172512351933
x20 = 25.9181393921158
x21 = 3.92699081698724
x22 = 96.6039740978861
x23 = -8.63937979737193
x24 = 91.8915851175014
x25 = -91.8915851175014
x26 = -13.3517687777566
x27 = 98.174770424681
x28 = -40.0553063332699
x29 = -96.6039740978861
x30 = -71.4712328691678
x31 = 77.7544181763474
x32 = -47.9092879672443
x33 = 33.7721210260903
x34 = 38.484510006475
x35 = -98.174770424681
x36 = -77.7544181763474
x37 = -57.3340659280137
x38 = 8.63937979737193
x39 = -27.4889357189107
x40 = 40.0553063332699
x41 = -52.621676947629
x42 = -69.9004365423729
x43 = 54.1924732744239
x44 = -2.35619449019234
x45 = -54.1924732744239
x46 = 90.3207887907066
x47 = -62.0464549083984
x48 = 69.9004365423729
x49 = 82.4668071567321
x50 = -74.6128255227576
x51 = -33.7721210260903
x52 = 74.6128255227576
x53 = -46.3384916404494
x54 = 62.0464549083984
x55 = 60.4756585816035
x56 = -68.329640215578
x57 = 85.6083998103219
x58 = -3.92699081698724
x59 = 10.2101761241668
x60 = 80.8960108299372
x61 = 35.3429173528852
x62 = 99.7455667514759
x63 = -63.6172512351933
x64 = -90.3207887907066
x65 = -30.6305283725005
x66 = -93.4623814442964
x67 = 19.6349540849362
x68 = 2.35619449019234
x69 = 11.7809724509617
x70 = -10.2101761241668
x71 = 18.0641577581413
x72 = 16.4933614313464
x73 = -24.3473430653209
x74 = -84.037603483527
x75 = -41.6261026600648
x76 = 30.6305283725005
x77 = 47.9092879672443
x78 = -35.3429173528852
x79 = -49.4800842940392
x80 = -11.7809724509617
x81 = 76.1836218495525
x82 = 41.6261026600648
x83 = -99.7455667514759
x83 = -99.7455667514759