Question

solve cosx.cos2x.cos3x = 1/4

Answer 1

cosx . cos2x . cos3x = 1/4
<=> cosx . cos3x . cos2x = 1/4
<=> 1/2.( cos4x + cos2x ).cos2x = 1/4
<=> ( cos4x + cos2x ).cos2x = 1/2
<=> cos4x.cos2x + cos²2x = 1/2
<=> (2cos²2x - 1)cos2x + cos²2x = 1/2
<=> 2cos^3(2x) - cos2x + cos²2x = 1/2
<=> 2cos^3(2x) + cos²2x - cos2x - 1/2 = 0
<=> 4cos^3(2x) + 2cos²2x - 2cos2x - 1 = 0
<=> 2cos²2x( 2cos2x + 1 ) - (2cos2x + 1 ) = 0
<=> ( 2cos2x + 1 )( 2cos²2x - 1 ) = 0

<=> 2cos2x + 1 = 0
<=> 2cos²2x - 1 = 0

<=> cos2x = -1/2
<=> cos²2x = 1/2

<=> cos2x = -1/2
<=> cos2x = √2/2
<=> cos2x = -√2/2

<=> 2x = 2π/3 + k2π ( k of Z )
<=> 2x = -2π/3 + k2π
<=> 2x = π/4 + k2π
<=> 2x = -π/4 + k2π
<=> 2x = 3π/4 + k2π
<=> 2x = -3π/4 + k2π

<=> x = π/3 + kπ
<=> x = -π/3 + kπ
<=> x = π/8 + kπ
<=> x = -π/8 + kπ
<=> x = 3π/8 + kπ
<=> x = -3π/8 + kπ

* * * * * * * * * * * * * * * * * * * *
cosx = 0

<=> x = π/2 + kπ ( k of Z )
<=> x = π/2 + k2π/2
<=> x = (1 + 2k)(π/2)