CPhill

If I understand this.....you  are  looking for   " - - "   ????

Looking at the right side

(1/2) [ sin pi/2 - sin pi/6 ]  =

(1/2) [ 1 - 1/2]  =

(1/2) (1/2)  =   1/4

So

" - - "  pi/3 * sin pi/6   = 1/4

" - - " pi/3 * (1/2) = 1/4           multiply both sides by 2

" - - " (pi/3)  =   1/2

The " - - " is    "cos "

2x > 3x - 3  ⇒  2x + 3 > 3x

3x - a > -6 ⇒   3x > a - 6

So....we have the combined inequality.....

a - 6 < 3x < 2x + 3          and when x  = 2

a - 6 < 3(2) < 2(2) + 3          

a - 6 < 6 < 7          add 6 to each term

a < 12 < 13

So  a  =  [ 1, 11] =   11 positive integers

We don't need the "squared identities," here...the double-angle identities serve better....

sin^2xcos^2x  = 

(sinx cosx)^2  =                                                                                

[  (1/2) sin 2x] ^2  =                                                         

 (1/4) (sin 2x)^2                       

Note, Cupcake.......     cos 4x  =   1 - 2[sin 2x]^2

                                    2[sin 2x ]^2  = 1 - cos(4x)

                                     (sin 2x)^2  =  [ 1 - cos(4x)] / 2   =  (1/2)[ 1 - cos (4x) ]

So we have

(1/4) (1/2)  (1 - cos(4x) ]    =

(1/8) [ 1 - cos (4x) ]  =

[ 1 - cos (4x) ]

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         8

Last one

6.   2.4^(3x+1) = 9    take the log of  both sides

log (2.4) ^(3x + 1)  = log 9      and we can write

(3x + 1) * log 2.4  = log 9

3x * log 2.4 + 1 * log 2.4 =  log 9

(3 log 2.4)x = log 9 - log 2.4     divide both sides by 3 log 2.4

x  =  [ log 9 - log 2.4 ] / [ 3* log 2.4 ] ≈  0.5033

4.    4^x  = 10

Take the log of each side

log 4^x  = log 10       and we can write

x * log 4  = 1      divide both sides by log 4

x  =  1 / log 4

5. (1/4)^2x = (1/2)^x

[ (1/2)^2 ]^(2x)  = (1/2)^x

(1/2)^(4x)  = (1/2)^x

4x  = x

4x - x  = 0

3x  = 0

x  = 0

1. 4^2x = 32^1/2  ......the idea is to get the bases the same and then solve for the exponents

(2^2) ^(2x)   = (2^5)^(1/2)

(2)^(4x)  = (2)^(5/2)

4x  = 5/2      divide both sides by 4

x  = 5/8

2. 9^x = 3^x-2

(3^2)^x  =3^(x - 2)

(3)^(2x)  = 3^(x - 2)

2x  = x - 2     subtract x from both sides

x = - 2

3. 2^x = 4^x+1

2^x  = (2^2)^(x + 1)

2^x = (2)^(2x + 2)

x = 2x + 2      subtract 2x from both sides

-1x  = 2         divide both sides by -1

x = -2

OK...

1. Slope at any point ...using implicit differentiation....  2x - 2y - 2xy ' + 8y y '  = 0

y' ( 8y - 2x ] = 2y - 2x

y '  =   [ 2y - 2x ] / [ 8y - 2x ]

2.  At x  = 2....we can find y as

2^2 - 2(2)y  + 4y^2  = 4

4 - 4y + 4y^2  = 84

4y^2 - 4y - 80 = 0

y^2 - y - 20  = 0       factor

(y - 5) (y + 4)  = 0     .....so   y = 5  or y = -4

So....we have the points   ( 2, 5) and (2, -4)

So  the slope at  (2, 5)  =  [2(5) - 2(2) ] / [ 8(5) - 2(2) ]  =   6/36  = 1/6

So...the equation is  y  = (1/6) (x - 2) + 5   

The slope at (2, -4) = [ 2(-4) - 2(2)] / [ 8(-4) -2(2)]  =  -12/ -36  =  1/3

So...the equation is  y  = (1/3)(x - 2) - 4

See the graph here : https://www.desmos.com/calculator/q55m709tum

I don't know the answer to the last one.....sorry  !!!

This one is a little tricky, Cupcake   !!!!

Not really kosher, but since we assume that 1 - tan^2x  is not equal to 0, we can multiply both sides by 1 - tan^2x  and get that

2tanx  = 1 - tan^2 x      rearrange as

tan^2 x + 2 tanx  - 1  = 0       let   tan x  =  m     and we have that

m^2 + 2m - 1  = 0

m^2 + 2m  =  1      complete the square on m

m^2 + 2m + 1  = 1 + 1

(m + 1)^2  = 2         take both roots

m + 1  = ±√2

m = ±√2 - 1

So  

tan x  = √2 - 1.....take the arctan (√2 - 1)  = m =   22.5°  =  pi/8 + n*pi

And

tan x  =  - √2 - 1.....take the arctan  (-√2 - 1 )  = x  = -67.5°.....add 180 to this  = 112.5°  = [225/2]°  = (5/4)pi / 2  =

(5/8)pi + n * pi

We are using the fact that   both functions are symmetric about the y axis

So

2                                                          2

 ∫ 12 - 3x^2   dx  is  the same  as    2 * ∫  12 - 3x^2  dx   

-2                                                         0