Mister Exam
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- Equation:
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Equation 14/(x-3)=-7/9
-
Equation x^4+2*x^2-8=0
-
Equation 9x−9(x+1)=−2x
-
Equation x^4+4=0
- Express {x} in terms of y where:
- -17*x+1*y=-11
- 1*x+4*y=20
- -18*x-2*y=9
- -20*x-14*y=5
-
Identical expressions
- two *sin(x/ three -pi/ four)= three
- 2 multiply by sinus of (x divide by 3 minus Pi divide by 4) equally 3
- two multiply by sinus of (x divide by three minus Pi divide by four) equally three
- 2sin(x/3-pi/4)=3
- 2sinx/3-pi/4=3
- 2*sin(x divide by 3-pi divide by 4)=3
-
Similar expressions
- 2*sin(x/3+pi/4)=3
2*sin(x/3-pi/4)=3 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
/x pi\
2*sin|- - --| = 3
\3 4 /
$$2 \sin{\left(\frac{x}{3} - \frac{\pi}{4} \right)} = 3$$
Detail solution
Given the equation
$$2 \sin{\left(\frac{x}{3} - \frac{\pi}{4} \right)} = 3$$
- this is the simplest trigonometric equation
The equation is transformed to
$$\cos{\left(\frac{x}{3} + \frac{\pi}{4} \right)} = - \frac{3}{2}$$
As right part of the equation
modulo =
but cos
can no be more than 1 or less than -1
so the solution of the equation d'not exist.
$$2 \sin{\left(\frac{x}{3} - \frac{\pi}{4} \right)} = 3$$
- this is the simplest trigonometric equation
Divide both parts of the equation by -2
The equation is transformed to
$$\cos{\left(\frac{x}{3} + \frac{\pi}{4} \right)} = - \frac{3}{2}$$
As right part of the equation
modulo =
True
but cos
can no be more than 1 or less than -1
so the solution of the equation d'not exist.
Rapid solution
[src]
3*pi
x1 = 3*re(acos(-3/2)) - ---- + 3*I*im(acos(-3/2))
4
$$x_{1} = - \frac{3 \pi}{4} + 3 \operatorname{re}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)} + 3 i \operatorname{im}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)}$$
21*pi
x2 = -3*re(acos(-3/2)) + ----- - 3*I*im(acos(-3/2))
4
$$x_{2} = - 3 \operatorname{re}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)} + \frac{21 \pi}{4} - 3 i \operatorname{im}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)}$$
x2 = -3*re(acos(-3/2)) + 21*pi/4 - 3*i*im(acos(-3/2))
Sum and product of roots
[src]
sum
3*pi 21*pi
3*re(acos(-3/2)) - ---- + 3*I*im(acos(-3/2)) + -3*re(acos(-3/2)) + ----- - 3*I*im(acos(-3/2))
4 4
$$\left(- \frac{3 \pi}{4} + 3 \operatorname{re}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)} + 3 i \operatorname{im}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)}\right) + \left(- 3 \operatorname{re}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)} + \frac{21 \pi}{4} - 3 i \operatorname{im}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)}\right)$$
=
9*pi ---- 2
$$\frac{9 \pi}{2}$$
product
/ 3*pi \ / 21*pi \ |3*re(acos(-3/2)) - ---- + 3*I*im(acos(-3/2))|*|-3*re(acos(-3/2)) + ----- - 3*I*im(acos(-3/2))| \ 4 / \ 4 /
$$\left(- \frac{3 \pi}{4} + 3 \operatorname{re}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)} + 3 i \operatorname{im}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)}\right) \left(- 3 \operatorname{re}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)} + \frac{21 \pi}{4} - 3 i \operatorname{im}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)}\right)$$
=
-9*(-pi + 4*re(acos(-3/2)) + 4*I*im(acos(-3/2)))*(-7*pi + 4*re(acos(-3/2)) + 4*I*im(acos(-3/2)))
------------------------------------------------------------------------------------------------
16
$$- \frac{9 \left(- 7 \pi + 4 \operatorname{re}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)} + 4 i \operatorname{im}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)}\right) \left(- \pi + 4 \operatorname{re}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)} + 4 i \operatorname{im}{\left(\operatorname{acos}{\left(- \frac{3}{2} \right)}\right)}\right)}{16}$$
-9*(-pi + 4*re(acos(-3/2)) + 4*i*im(acos(-3/2)))*(-7*pi + 4*re(acos(-3/2)) + 4*i*im(acos(-3/2)))/16
Numerical answer
[src]
x1 = 7.06858347057703 - 2.88727095035762*i
x2 = 7.06858347057703 + 2.88727095035762*i
x2 = 7.06858347057703 + 2.88727095035762*i