Question

solution of equation sin6x+cos4x +2=0​

Answer 1

Answer:

The solution is

$x=\frac{\pi}{4}$

Step-by-step explanation:

Formula used:

$sin3A= 3 sinA - 4 sin^3A\\\\cos2A=1-2sin^2A$

sin6x + cos4x +2 =0

sin3(2x)+cos2(2x)+2=0

$3 sin2x - 4 sin^3(2x)+1-2sin^2(2x)+2=0$

$3 sin2x - 4sin^3(2x)-2sin^2(2x)+3=0$

$3sin(2x) - 4sin^3(2x)-2sin^2(2x)+3=0$

$4sin^3(2x)+2sin^2(2x)-3sin(2x)-3=0$

$4t^3+2t^2-3t-3=0$

where t = sin(2x)

since the sum of the coefficients is zero, then (t-1) is a factor

1  | 4    2     -3    -3

  |       4      6      3

  |_________________

    4     6     3     | 0

                         -------------

Now

$4t^3+2t^2-3t-3=(t-1)(4t^2+6t+3)$

$4t^3+2t^2-3t-3=0$

$(t-1)(4t^2+6t+3)=0$

$4t^2+6t+3=0$

It gives imaginary values for t

t -1 =0

t = 1

sin(2x) = 1

$2x=\frac{\pi}{2}$

$x=\frac{\pi}{4}$

The solution is

$x=\frac{\pi}{4}$