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- La ecuación:
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Ecuación 2x^2-x-1=x^2-5x-(-1-x^2)
-
Ecuación x(x^2+2x+1)=6(x+1)
-
Ecuación 4x+1=-3x+15
- Ecuación (3x−36)⋅(x+8)=0
- Expresar {x} en función de y en la ecuación:
- -6*x-20*y=8
- 13*x-12*y=19
- -9*x+11*y=9
- -4*x-8*y=-20
- Derivada de:
-
cost
-
2cos2t
- Integral de d{x}:
- cost
- 2cos2t
-
Expresiones idénticas
- cost=2cos2t
- coseno de t es igual a 2 coseno de 2t
-
Expresiones semejantes
- y=2*t*sin(t)-(t^2-2)*cos(t)
- (cos(t)+2)*3*k-2*sin(t)=0
-
Expresiones con funciones
- cost
- cost=3,4
- cost=g-(c+d)-f-(g*e)
cost=2cos2t la ecuación
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
Tenemos la ecuación
$$\cos{\left(t \right)} = 2 \cos{\left(2 t \right)}$$
cambiamos
$$- 4 \cos^{2}{\left(t \right)} + \cos{\left(t \right)} - 1 = 0$$
$$- 4 \cos^{2}{\left(t \right)} + \cos{\left(t \right)} - 1 = 0$$
Sustituimos
$$w = \cos{\left(t \right)}$$
Es la ecuación de la forma
La ecuación cuadrática puede ser resuelta
con la ayuda del discriminante.
Las raíces de la ecuación cuadrática:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
donde D = b^2 - 4*a*c es el discriminante.
Como
$$a = -4$$
$$b = 1$$
$$c = -1$$
, entonces
Como D < 0 la ecuación
no tiene raíces reales,
pero hay raíces complejas.
o
$$w_{1} = \frac{1}{8} - \frac{\sqrt{15} i}{8}$$
$$w_{2} = \frac{1}{8} + \frac{\sqrt{15} i}{8}$$
hacemos cambio inverso
$$\cos{\left(t \right)} = w$$
Tenemos la ecuación
$$\cos{\left(t \right)} = w$$
es la ecuación trigonométrica más simple
Esta ecuación se reorganiza en
$$t = \pi n + \operatorname{acos}{\left(w \right)}$$
$$t = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
O
$$t = \pi n + \operatorname{acos}{\left(w \right)}$$
$$t = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, donde n es cualquier número entero
sustituimos w:
$$t_{1} = \pi n + \operatorname{acos}{\left(w_{1} \right)}$$
$$t_{1} = \pi n + \operatorname{acos}{\left(\frac{1}{8} - \frac{\sqrt{15} i}{8} \right)}$$
$$t_{1} = \pi n + \operatorname{acos}{\left(\frac{1}{8} - \frac{\sqrt{15} i}{8} \right)}$$
$$t_{2} = \pi n + \operatorname{acos}{\left(w_{2} \right)}$$
$$t_{2} = \pi n + \operatorname{acos}{\left(\frac{1}{8} + \frac{\sqrt{15} i}{8} \right)}$$
$$t_{2} = \pi n + \operatorname{acos}{\left(\frac{1}{8} + \frac{\sqrt{15} i}{8} \right)}$$
$$t_{3} = \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi$$
$$t_{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{8} - \frac{\sqrt{15} i}{8} \right)}$$
$$t_{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{8} - \frac{\sqrt{15} i}{8} \right)}$$
$$t_{4} = \pi n + \operatorname{acos}{\left(w_{2} \right)} - \pi$$
$$t_{4} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{8} + \frac{\sqrt{15} i}{8} \right)}$$
$$t_{4} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{8} + \frac{\sqrt{15} i}{8} \right)}$$
$$\cos{\left(t \right)} = 2 \cos{\left(2 t \right)}$$
cambiamos
$$- 4 \cos^{2}{\left(t \right)} + \cos{\left(t \right)} - 1 = 0$$
$$- 4 \cos^{2}{\left(t \right)} + \cos{\left(t \right)} - 1 = 0$$
Sustituimos
$$w = \cos{\left(t \right)}$$
Es la ecuación de la forma
a*w^2 + b*w + c = 0
La ecuación cuadrática puede ser resuelta
con la ayuda del discriminante.
Las raíces de la ecuación cuadrática:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
donde D = b^2 - 4*a*c es el discriminante.
Como
$$a = -4$$
$$b = 1$$
$$c = -1$$
, entonces
D = b^2 - 4 * a * c =
(1)^2 - 4 * (-4) * (-1) = -15
Como D < 0 la ecuación
no tiene raíces reales,
pero hay raíces complejas.
w1 = (-b + sqrt(D)) / (2*a)
w2 = (-b - sqrt(D)) / (2*a)
o
$$w_{1} = \frac{1}{8} - \frac{\sqrt{15} i}{8}$$
$$w_{2} = \frac{1}{8} + \frac{\sqrt{15} i}{8}$$
hacemos cambio inverso
$$\cos{\left(t \right)} = w$$
Tenemos la ecuación
$$\cos{\left(t \right)} = w$$
es la ecuación trigonométrica más simple
Esta ecuación se reorganiza en
$$t = \pi n + \operatorname{acos}{\left(w \right)}$$
$$t = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
O
$$t = \pi n + \operatorname{acos}{\left(w \right)}$$
$$t = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, donde n es cualquier número entero
sustituimos w:
$$t_{1} = \pi n + \operatorname{acos}{\left(w_{1} \right)}$$
$$t_{1} = \pi n + \operatorname{acos}{\left(\frac{1}{8} - \frac{\sqrt{15} i}{8} \right)}$$
$$t_{1} = \pi n + \operatorname{acos}{\left(\frac{1}{8} - \frac{\sqrt{15} i}{8} \right)}$$
$$t_{2} = \pi n + \operatorname{acos}{\left(w_{2} \right)}$$
$$t_{2} = \pi n + \operatorname{acos}{\left(\frac{1}{8} + \frac{\sqrt{15} i}{8} \right)}$$
$$t_{2} = \pi n + \operatorname{acos}{\left(\frac{1}{8} + \frac{\sqrt{15} i}{8} \right)}$$
$$t_{3} = \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi$$
$$t_{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{8} - \frac{\sqrt{15} i}{8} \right)}$$
$$t_{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{8} - \frac{\sqrt{15} i}{8} \right)}$$
$$t_{4} = \pi n + \operatorname{acos}{\left(w_{2} \right)} - \pi$$
$$t_{4} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{8} + \frac{\sqrt{15} i}{8} \right)}$$
$$t_{4} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{8} + \frac{\sqrt{15} i}{8} \right)}$$
Respuesta rápida
[src]
/ _____________\
| ____ ___ / ____ |
|1 \/ 33 I*\/ 2 *\/ 15 - \/ 33 |
t1 = -I*log|- + ------ - ------------------------|
\8 8 8 /
$$t_{1} = - i \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} - \frac{\sqrt{2} i \sqrt{15 - \sqrt{33}}}{8} \right)}$$
/ _____________\
| ____ ___ / ____ |
|1 \/ 33 I*\/ 2 *\/ 15 - \/ 33 |
t2 = -I*log|- + ------ + ------------------------|
\8 8 8 /
$$t_{2} = - i \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} + \frac{\sqrt{2} i \sqrt{15 - \sqrt{33}}}{8} \right)}$$
/ _____________\
| ____ ___ / ____ |
|1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 |
t3 = -I*log|- - ------ - ------------------------|
\8 8 8 /
$$t_{3} = - i \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} - \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)}$$
/ _____________\
| ____ ___ / ____ |
|1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 |
t4 = -I*log|- - ------ + ------------------------|
\8 8 8 /
$$t_{4} = - i \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} + \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)}$$
t4 = -i*log(-sqrt(33)/8 + 1/8 + sqrt(2)*i*sqrt(sqrt(33) + 15)/8)
Suma y producto de raíces
[src]
suma
/ _____________\ / _____________\ / _____________\ / _____________\
| ____ ___ / ____ | | ____ ___ / ____ | | ____ ___ / ____ | | ____ ___ / ____ |
|1 \/ 33 I*\/ 2 *\/ 15 - \/ 33 | |1 \/ 33 I*\/ 2 *\/ 15 - \/ 33 | |1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 | |1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 |
- I*log|- + ------ - ------------------------| - I*log|- + ------ + ------------------------| - I*log|- - ------ - ------------------------| - I*log|- - ------ + ------------------------|
\8 8 8 / \8 8 8 / \8 8 8 / \8 8 8 /
$$\left(- i \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} - \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)} + \left(- i \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} - \frac{\sqrt{2} i \sqrt{15 - \sqrt{33}}}{8} \right)} - i \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} + \frac{\sqrt{2} i \sqrt{15 - \sqrt{33}}}{8} \right)}\right)\right) - i \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} + \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)}$$
=
/ _____________\ / _____________\ / _____________\ / _____________\
| ____ ___ / ____ | | ____ ___ / ____ | | ____ ___ / ____ | | ____ ___ / ____ |
|1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 | |1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 | |1 \/ 33 I*\/ 2 *\/ 15 - \/ 33 | |1 \/ 33 I*\/ 2 *\/ 15 - \/ 33 |
- I*log|- - ------ - ------------------------| - I*log|- - ------ + ------------------------| - I*log|- + ------ - ------------------------| - I*log|- + ------ + ------------------------|
\8 8 8 / \8 8 8 / \8 8 8 / \8 8 8 /
$$- i \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} - \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)} - i \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} - \frac{\sqrt{2} i \sqrt{15 - \sqrt{33}}}{8} \right)} - i \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} + \frac{\sqrt{2} i \sqrt{15 - \sqrt{33}}}{8} \right)} - i \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} + \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)}$$
producto
/ _____________\ / / _____________\\ / / _____________\\ / / _____________\\
| ____ ___ / ____ | | | ____ ___ / ____ || | | ____ ___ / ____ || | | ____ ___ / ____ ||
|1 \/ 33 I*\/ 2 *\/ 15 - \/ 33 | | |1 \/ 33 I*\/ 2 *\/ 15 - \/ 33 || | |1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 || | |1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 ||
-I*log|- + ------ - ------------------------|*|-I*log|- + ------ + ------------------------||*|-I*log|- - ------ - ------------------------||*|-I*log|- - ------ + ------------------------||
\8 8 8 / \ \8 8 8 // \ \8 8 8 // \ \8 8 8 //
$$- i \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} + \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)} - i \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} - \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)} - i \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} - \frac{\sqrt{2} i \sqrt{15 - \sqrt{33}}}{8} \right)} \left(- i \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} + \frac{\sqrt{2} i \sqrt{15 - \sqrt{33}}}{8} \right)}\right)$$
=
/ ______________\ / _____________\ / ______________\ / ______________\ | ____ ___ / ____ | | ____ ___ / ____ | | ____ ___ / ____ | | ____ ___ / ____ | |1 \/ 33 \/ 2 *\/ -15 - \/ 33 | |1 \/ 33 I*\/ 2 *\/ 15 + \/ 33 | |1 \/ 33 \/ 2 *\/ -15 + \/ 33 | |1 \/ 33 \/ 2 *\/ -15 + \/ 33 | log|- - ------ + -----------------------|*log|- - ------ - ------------------------|*log|- + ------ - -----------------------|*log|- + ------ + -----------------------| \8 8 8 / \8 8 8 / \8 8 8 / \8 8 8 /
$$\log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} - \frac{\sqrt{2} \sqrt{-15 + \sqrt{33}}}{8} \right)} \log{\left(\frac{1}{8} + \frac{\sqrt{33}}{8} + \frac{\sqrt{2} \sqrt{-15 + \sqrt{33}}}{8} \right)} \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} + \frac{\sqrt{2} \sqrt{-15 - \sqrt{33}}}{8} \right)} \log{\left(- \frac{\sqrt{33}}{8} + \frac{1}{8} - \frac{\sqrt{2} i \sqrt{\sqrt{33} + 15}}{8} \right)}$$
log(1/8 - sqrt(33)/8 + sqrt(2)*sqrt(-15 - sqrt(33))/8)*log(1/8 - sqrt(33)/8 - i*sqrt(2)*sqrt(15 + sqrt(33))/8)*log(1/8 + sqrt(33)/8 - sqrt(2)*sqrt(-15 + sqrt(33))/8)*log(1/8 + sqrt(33)/8 + sqrt(2)*sqrt(-15 + sqrt(33))/8)
Respuesta numérica
[src]
t1 = -46.1879603481856
t2 = 68.5472090060468
t3 = -77.6038868840835
t4 = 18.2817265486101
t5 = -52.4711456553652
t6 = -96.4534428056223
t7 = 41.7766339523286
t8 = 0.567829372928644
t9 = 77.6038868840835
t10 = -4.07752210925112
t11 = 99.9631355419447
t12 = -38.2669412160062
t13 = 58.7543309625447
t14 = -5.71535593425094
t15 = 88.5324236734428
t16 = -35.4934486451491
t17 = -27.3384044266468
t18 = -69.6828677519041
t19 = -75.9660530590837
t20 = 62.2640236988672
t21 = 46.1879603481856
t22 = -43.4144677773285
t23 = -18.2817265486101
t24 = 35.4934486451491
t25 = 24.5649118557897
t26 = 8.48884850510805
t27 = 90.1702574984427
t28 = -11.9985412414305
t29 = -79.4757457954062
t30 = -92.0421164097653
t31 = 31.9837559088266
t32 = 25.700570601647
t33 = -39.904775041006
t34 = -82.2492383662633
t35 = -54.3430045666878
t36 = -25.700570601647
t37 = -31.9837559088266
t38 = -85.7589311025857
t39 = 81.113579620406
t40 = 21.0552191194672
t41 = -8.48884850510805
t42 = 83.8870721912631
t43 = 48.0598192595082
t44 = 16.6438927236103
t45 = -41.7766339523286
t46 = 11.9985412414305
t47 = -10.3607074164307
t48 = 69.6828677519041
t49 = 96.4534428056223
t50 = 38.2669412160062
t51 = -2.20566319792847
t52 = -63.3996824447245
t53 = -99.9631355419447
t54 = 5.71535593425094
t55 = 39.904775041006
t56 = 52.4711456553652
t57 = 93.6799502347652
t58 = 75.9660530590837
t59 = 2.20566319792847
t60 = 10.3607074164307
t61 = 82.2492383662633
t62 = 85.7589311025857
t63 = -55.9808383916876
t64 = -87.3967649275856
t65 = 92.0421164097653
t66 = 55.9808383916876
t67 = 44.5501265231857
t68 = 54.3430045666878
t69 = 33.6215897338264
t70 = -24.5649118557897
t71 = -83.8870721912631
t72 = 60.6261898738674
t73 = 49.697653084508
t74 = -62.2640236988672
t75 = -68.5472090060468
t76 = -98.3253017169449
t77 = 98.3253017169449
t78 = -48.0598192595082
t79 = -71.3207015769039
t80 = -49.697653084508
t81 = -33.6215897338264
t82 = 4.07752210925112
t83 = -60.6261898738674
t84 = -19.4173852944674
t85 = -93.6799502347652
t86 = -90.1702574984427
t86 = -90.1702574984427