If it's not what You are looking for type in the equation solver your own equation and let us solve it.
D( x )
cos(x) = 0
cos(x) = 0
cos(x) = 0
cos(x) = 0 <=> x = pi*k_1+pi/2 i k_1 należy do I
t_1 = pi*k_1+pi/2
x-t_1 = 0
x-t_1 = 0 // + t_1
x = t_1
x = pi*k_1+pi/2 i k_1 należy do I
x in {( -oo : +oo ) / < pi*k_1+pi/2 : pi*k_1+pi/2 >} i k_1 -> {I}
(2*cos(3*x))/cos(x) = 0
(2*cos(3*x))/cos(x) = 0 // * cos(x)
2*cos(3*x) = 0
cos(3*x) = 0
cos(3*x) = 0 <=> 3*x = pi*k_1+pi/2 i k_1 należy do I
t_1 = pi*k_1+pi/2
3*x-t_1 = 0
3*x-t_1 = 0 // + t_1
3*x = t_1 // : 3
x = t_1/3
x = pi*k_1+pi/2/3 i k_1 należy do I
2*cos(3*x) = 0 <=> 2 = 0 or 2*cos(3*x) = 0 <=> cos(3*x) = 0
x = pi*k_1+pi/2/3
| log^7(12x)=2+log^7(4) | | 11-(16-5x)=6(7+2x) | | X=22-y | | 0.2c=4 | | 15x+5000=40x | | -5x=8*6+79 | | 15x+500=35x | | 15x+500=35 | | a^2+2a^2=24 | | 0.6d+3.4=16 | | F(x)=3x—9/2 | | logx^40=-2 | | (6x-4)+(5x-8)= | | 2y+4=y-24 | | 0.25/50 | | 2log(-2x)=0 | | (5x-3)+(2x-1)= | | s*7-42=133 | | z/10-2=5 | | (5x-3)*(2x-1)= | | .40x+.20(180)=.20(150) | | x+x+x=x+10+2 | | x+x+10-10+2x-a=180+3x | | 2x-36= | | 1u(1u-5)+8u=1u(1u+2)-4 | | 7a(2a-1)=0 | | x+x+10-10+2x-a=180 | | t/3-7=16 | | 3a+a=8+a | | (23+x)*x=(12x+7)*2x | | 1t(1t+4)-1=1t(1t+2)+2 | | 2.2=3x-3.8 |