Answer:

LET x = 18°

so, 5x = 90°

now we can write

2x + 3x = 90°

so 2x = 90° - 3x

now taking sin both SIDE we can write

sin2x = sin(90°-3x)

sin2x = cos3x [as we know, sin(90°-3x) = Cos3x ]

so evaluating we can write

2sinxcosx =4cos³x - 3cosx [as we know, Cos3x = 4cos³x - 3cosx, This I will explain later ]

Now, 2sinxcosx - 4cos³x + 3cosx = 0

cosx(2sinx - 4cos²x + 3) = 0

Now divding both side by cosx we get,

2sinx - 4cos²x + 3 = 0

2sinx - 4(1-sin²x) + 3 = 0 [as we know, cos²x = (1-sin²x), by sin²x + cos²x = 1 ]

2sinx - 4 + 4sin²x + 3 = 0

2sinx + 4sin²x - 1 = 0

we can write it as,

4sin²x + 2sinx - 1 = 0

Now apply Sridhar Acharya Formula Here,

ax² + bx + C = 0

so, x = (-b ± √(b² - 4ac))/2a

now applying it in the equation

SINX =

(-2 ± √(2² - 44(-1)))/2.(4)

sinx = (-2 ± √(4 +16))/8

sinx = (-2 ± √20)/8

sinx = (-2 ± 2.√5) / 8

sin x = 2(-1 ± √5 ) / 8

sin x = (-1 ± √5)/4

sin18° = (-1 ± √5)/4

Now how I get the Cos3x value?

Cox3x = Cos(2x+x)

Cox3x = Cos2xCosx - Sin2xSinx

Cox3x = (2Cos²x - 1)Cosx - (2SinxCosx)Sinx

Cox3x = 2Cos³x -Cosx - 2Sin²xCosx

Cox3x = 2Cos³x -Cosx - 2Cosx(1-Cos²x)

Cox3x = 2Cos³x -Cosx - 2Cosx + 2Cos³x

Cox3x = 4Cos³x - 3Cosx