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2x+10=12 2x=2 x=1
2 plus 1-x divided by.
8x-4
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
2x + 7 = 10 2x + 7 - 7 = 10 - 7 2x = 3 2x/2 = 3/2 x = 1 1/2 OR 1.5 The answer is 1 1/2, also written as 1.5
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
2x+10=12 2x=2 x=1
7 plus 7x--2x plus 3 is: 7 + 7x + 2x + 3 = 0 9x + 10 = 0 9x = -10 x = -10/9 = -1 1/9
2 plus 1-x divided by.
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
16 + 2x = 10 + 8x 2x - 8x = 10 - 16 -6x = -6 x = 1 Check: 16 + 2(1) = 10 + 8(1) 18 = 18
8x-4
10 + 2(x - 1) + 7 x 3x + 1 =10 + 2x - 2 + (7 x 3x) + 1 =10 + 2x - 2 + 21x + 1 =2x + 21x + 10 - 2 + 1 =23x + 9
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
2x + 7 = 10 2x + 7 - 7 = 10 - 7 2x = 3 2x/2 = 3/2 x = 1 1/2 OR 1.5 The answer is 1 1/2, also written as 1.5
I'm answering this based on the way you wrote it. If you meant something different, please substitute words where necessary. x2-2x+4+2x+1-x62+5=2x-2x+4+2x+1-62x+5=-60x+10
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x