6sin^2x–11cosx–10=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
( 6 sin 2 ( x ) − 11 cos ( x ) ) − 10 = 0 \left(6 \sin^{2}{\left(x \right)} - 11 \cos{\left(x \right)}\right) - 10 = 0 ( 6 sin 2 ( x ) − 11 cos ( x ) ) − 10 = 0 transform
6 sin 2 ( x ) − 11 cos ( x ) − 10 = 0 6 \sin^{2}{\left(x \right)} - 11 \cos{\left(x \right)} - 10 = 0 6 sin 2 ( x ) − 11 cos ( x ) − 10 = 0 − 6 cos 2 ( x ) − 11 cos ( x ) − 4 = 0 - 6 \cos^{2}{\left(x \right)} - 11 \cos{\left(x \right)} - 4 = 0 − 6 cos 2 ( x ) − 11 cos ( x ) − 4 = 0 Do replacement
w = cos ( x ) w = \cos{\left(x \right)} w = cos ( x ) 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 = D − b 2 a w_{1} = \frac{\sqrt{D} - b}{2 a} w 1 = 2 a D − b w 2 = − D − b 2 a w_{2} = \frac{- \sqrt{D} - b}{2 a} w 2 = 2 a − D − b where D = b^2 - 4*a*c - it is the discriminant.
Because
a = − 6 a = -6 a = − 6 b = − 11 b = -11 b = − 11 c = − 4 c = -4 c = − 4 , then
D = b^2 - 4 * a * c = (-11)^2 - 4 * (-6) * (-4) = 25 Because D > 0, then the equation has two roots.
w1 = (-b + sqrt(D)) / (2*a) w2 = (-b - sqrt(D)) / (2*a) or
w 1 = − 4 3 w_{1} = - \frac{4}{3} w 1 = − 3 4 w 2 = − 1 2 w_{2} = - \frac{1}{2} w 2 = − 2 1 do backward replacement
cos ( x ) = w \cos{\left(x \right)} = w cos ( x ) = w Given the equation
cos ( x ) = w \cos{\left(x \right)} = w cos ( x ) = w - this is the simplest trigonometric equation
This equation is transformed to
x = π n + acos ( w ) x = \pi n + \operatorname{acos}{\left(w \right)} x = πn + acos ( w ) x = π n + acos ( w ) − π x = \pi n + \operatorname{acos}{\left(w \right)} - \pi x = πn + acos ( w ) − π Or
x = π n + acos ( w ) x = \pi n + \operatorname{acos}{\left(w \right)} x = πn + acos ( w ) x = π n + acos ( w ) − π x = \pi n + \operatorname{acos}{\left(w \right)} - \pi x = πn + acos ( w ) − π , where n - is a integer
substitute w:
x 1 = π n + acos ( w 1 ) x_{1} = \pi n + \operatorname{acos}{\left(w_{1} \right)} x 1 = πn + acos ( w 1 ) x 1 = π n + acos ( − 4 3 ) x_{1} = \pi n + \operatorname{acos}{\left(- \frac{4}{3} \right)} x 1 = πn + acos ( − 3 4 ) x 1 = π n + acos ( − 4 3 ) x_{1} = \pi n + \operatorname{acos}{\left(- \frac{4}{3} \right)} x 1 = πn + acos ( − 3 4 ) x 2 = π n + acos ( w 2 ) x_{2} = \pi n + \operatorname{acos}{\left(w_{2} \right)} x 2 = πn + acos ( w 2 ) x 2 = π n + acos ( − 1 2 ) x_{2} = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)} x 2 = πn + acos ( − 2 1 ) x 2 = π n + 2 π 3 x_{2} = \pi n + \frac{2 \pi}{3} x 2 = πn + 3 2 π x 3 = π n + acos ( w 1 ) − π x_{3} = \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi x 3 = πn + acos ( w 1 ) − π x 3 = π n − π + acos ( − 4 3 ) x_{3} = \pi n - \pi + \operatorname{acos}{\left(- \frac{4}{3} \right)} x 3 = πn − π + acos ( − 3 4 ) x 3 = π n − π + acos ( − 4 3 ) x_{3} = \pi n - \pi + \operatorname{acos}{\left(- \frac{4}{3} \right)} x 3 = πn − π + acos ( − 3 4 ) x 4 = π n + acos ( w 2 ) − π x_{4} = \pi n + \operatorname{acos}{\left(w_{2} \right)} - \pi x 4 = πn + acos ( w 2 ) − π x 4 = π n − π + acos ( − 1 2 ) x_{4} = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)} x 4 = πn − π + acos ( − 2 1 ) x 4 = π n − π 3 x_{4} = \pi n - \frac{\pi}{3} x 4 = πn − 3 π
The graph
0 -80 -60 -40 -20 20 40 60 80 -100 100 -25 25
x 1 = − 2 π 3 x_{1} = - \frac{2 \pi}{3} x 1 = − 3 2 π
x 2 = 2 π 3 x_{2} = \frac{2 \pi}{3} x 2 = 3 2 π
/ / ___\\ / / ___\\
x3 = 2*im\atanh\\/ 7 // - 2*I*re\atanh\\/ 7 //
x 3 = 2 im ( atanh ( 7 ) ) − 2 i re ( atanh ( 7 ) ) x_{3} = 2 \operatorname{im}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} - 2 i \operatorname{re}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} x 3 = 2 im ( atanh ( 7 ) ) − 2 i re ( atanh ( 7 ) )
/ / ___\\ / / ___\\
x4 = - 2*im\atanh\\/ 7 // + 2*I*re\atanh\\/ 7 //
x 4 = − 2 im ( atanh ( 7 ) ) + 2 i re ( atanh ( 7 ) ) x_{4} = - 2 \operatorname{im}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} + 2 i \operatorname{re}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} x 4 = − 2 im ( atanh ( 7 ) ) + 2 i re ( atanh ( 7 ) )
x4 = -2*im(atanh(sqrt(7))) + 2*i*re(atanh(sqrt(7)))
Sum and product of roots
[src]
2*pi 2*pi / / ___\\ / / ___\\ / / ___\\ / / ___\\
- ---- + ---- + 2*im\atanh\\/ 7 // - 2*I*re\atanh\\/ 7 // + - 2*im\atanh\\/ 7 // + 2*I*re\atanh\\/ 7 //
3 3
( ( − 2 π 3 + 2 π 3 ) + ( 2 im ( atanh ( 7 ) ) − 2 i re ( atanh ( 7 ) ) ) ) + ( − 2 im ( atanh ( 7 ) ) + 2 i re ( atanh ( 7 ) ) ) \left(\left(- \frac{2 \pi}{3} + \frac{2 \pi}{3}\right) + \left(2 \operatorname{im}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} - 2 i \operatorname{re}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)}\right)\right) + \left(- 2 \operatorname{im}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} + 2 i \operatorname{re}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)}\right) ( ( − 3 2 π + 3 2 π ) + ( 2 im ( atanh ( 7 ) ) − 2 i re ( atanh ( 7 ) ) ) ) + ( − 2 im ( atanh ( 7 ) ) + 2 i re ( atanh ( 7 ) ) )
-2*pi 2*pi / / / ___\\ / / ___\\\ / / / ___\\ / / ___\\\
-----*----*\2*im\atanh\\/ 7 // - 2*I*re\atanh\\/ 7 ///*\- 2*im\atanh\\/ 7 // + 2*I*re\atanh\\/ 7 ///
3 3
− 2 π 3 2 π 3 ( 2 im ( atanh ( 7 ) ) − 2 i re ( atanh ( 7 ) ) ) ( − 2 im ( atanh ( 7 ) ) + 2 i re ( atanh ( 7 ) ) ) - \frac{2 \pi}{3} \frac{2 \pi}{3} \left(2 \operatorname{im}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} - 2 i \operatorname{re}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)}\right) \left(- 2 \operatorname{im}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} + 2 i \operatorname{re}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)}\right) − 3 2 π 3 2 π ( 2 im ( atanh ( 7 ) ) − 2 i re ( atanh ( 7 ) ) ) ( − 2 im ( atanh ( 7 ) ) + 2 i re ( atanh ( 7 ) ) )
2
2 / / / ___\\ / / ___\\\
16*pi *\- I*re\atanh\\/ 7 // + im\atanh\\/ 7 ///
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9
16 π 2 ( im ( atanh ( 7 ) ) − i re ( atanh ( 7 ) ) ) 2 9 \frac{16 \pi^{2} \left(\operatorname{im}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)} - i \operatorname{re}{\left(\operatorname{atanh}{\left(\sqrt{7} \right)}\right)}\right)^{2}}{9} 9 16 π 2 ( im ( atanh ( 7 ) ) − i re ( atanh ( 7 ) ) ) 2
16*pi^2*(-i*re(atanh(sqrt(7))) + im(atanh(sqrt(7))))^2/9