Mister Exam
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How to use it?
- Equation:
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Equation sin6x=1/2
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Equation tgx=3/2
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Equation sin((2*x+2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sin(x)^2/cos(-(-pi)/3)
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Equation tgx=-(3/4)
- Express {x} in terms of y where:
- 15*x-15*y=16
- 9*x+18*y=6
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
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Identical expressions
- sin((two *x+ two *pi)/ three)*cos((four *x+pi)/ three)-cos(two *x)=sin(x)^ two /cos(-(-pi)/ three)
- sinus of ((2 multiply by x plus 2 multiply by Pi ) divide by 3) multiply by co sinus of e of ((4 multiply by x plus Pi ) divide by 3) minus co sinus of e of (2 multiply by x) equally sinus of (x) squared divide by co sinus of e of ( minus ( minus Pi ) divide by 3)
- sinus of ((two multiply by x plus two multiply by Pi ) divide by three) multiply by co sinus of e of ((four multiply by x plus Pi ) divide by three) minus co sinus of e of (two multiply by x) equally sinus of (x) to the power of two divide by co sinus of e of ( minus ( minus Pi ) divide by three)
- sin((2*x+2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sin(x)2/cos(-(-pi)/3)
- sin2*x+2*pi/3*cos4*x+pi/3-cos2*x=sinx2/cos--pi/3
- sin((2*x+2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sin(x)²/cos(-(-pi)/3)
- sin((2*x+2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sin(x) to the power of 2/cos(-(-pi)/3)
- sin((2x+2pi)/3)cos((4x+pi)/3)-cos(2x)=sin(x)^2/cos(-(-pi)/3)
- sin((2x+2pi)/3)cos((4x+pi)/3)-cos(2x)=sin(x)2/cos(-(-pi)/3)
- sin2x+2pi/3cos4x+pi/3-cos2x=sinx2/cos--pi/3
- sin2x+2pi/3cos4x+pi/3-cos2x=sinx^2/cos--pi/3
- sin((2*x+2*pi) divide by 3)*cos((4*x+pi) divide by 3)-cos(2*x)=sin(x)^2 divide by cos(-(-pi) divide by 3)
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Similar expressions
- sin(2*x+2*pi/3)*cos((4*x+pi)/3)-cos(2*x)=sin(2*x)/cos(-(-pi)/3)
- sin((2*x+2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sin(x)^2/cos(-(pi)/3)
- sin((2*x+2*pi)/3)*cos((4*x+pi)/3)+cos(2*x)=sin(x)^2/cos(-(-pi)/3)
- sin((2*x-2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sin(x)^2/cos(-(-pi)/3)
- sin((2*x+2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sin(x)^2/cos((-pi)/3)
- sin((2*x+2*pi)/3)*cos((4*x-pi)/3)-cos(2*x)=sin(x)^2/cos(-(-pi)/3)
- sin((2*x+2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sinx^2/cos(-(-pi)/3)
sin((2*x+2*pi)/3)*cos((4*x+pi)/3)-cos(2*x)=sin(x)^2/cos(-(-pi)/3) equation
The teacher will be very surprised to see your correct solution 😉
The solution
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2
/2*x + 2*pi\ /4*x + pi\ sin (x)
sin|----------|*cos|--------| - cos(2*x) = -------
\ 3 / \ 3 / /pi\
cos|--|
\3 /
$$\sin{\left(\frac{2 x + 2 \pi}{3} \right)} \cos{\left(\frac{4 x + \pi}{3} \right)} - \cos{\left(2 x \right)} = \frac{\sin^{2}{\left(x \right)}}{\cos{\left(\frac{\pi}{3} \right)}}$$
The graph