What is the general solution for 2cos(a)sin(a)-4cos(a)+6sin(a)-12=0 ?

The general solution for 2cos(a)sin(a)-4cos(a)+6sin(a)-12=0 is No Solution for a\in\mathbb{R}

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2cos(a)sin(a)-4cos(a)+6sin(a)-12=0

2 cos (a) sin (a) - 4 cos (a) + 6 sin (a) - 12 = 0

No Solution for a \in R

Solution steps

2 cos (a) sin (a) - 4 cos (a) + 6 sin (a) - 12 = 0

Factor

2 cos (a) sin (a) - 4 cos (a) + 6 sin (a) - 12 : 2 (sin (a) - 2) (cos (a) + 3)

2 (sin (a) - 2) (cos (a) + 3) = 0

Solving each part separately

sin (a) - 2 = 0 or cos (a) + 3 = 0

sin (a) - 2 = 0 :

No Solution

cos (a) + 3 = 0 :

No Solution

Combine all the solutions

No Solution for a \in R

cos (x) = \frac{83}{92}

sin (\frac{π}{4} x) = 1

3 cos (x) + 1 = - 2 cos^{2} (x)

3 + 2 sin (β) = 0

sin (2 t) = \frac{3}{2}

What is the general solution for 2cos(a)sin(a)-4cos(a)+6sin(a)-12=0 ?
The general solution for 2cos(a)sin(a)-4cos(a)+6sin(a)-12=0 is No Solution for a\in\mathbb{R}