What is the general solution for 12cos(x)+6sin(2x)=0,0<= x<= 2pi ?

The general solution for 12cos(x)+6sin(2x)=0,0<= x<= 2pi is x= pi/2 ,x=(3pi)/2

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12cos(x)+6sin(2x)=0,0<= x<= 2pi

12 cos (x) + 6 sin (2 x) = 0, 0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

12 cos (x) + 6 sin (2 x) = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

12 cos (x) + 12 cos (x) sin (x) = 0

Factor

12 cos (x) + 12 cos (x) sin (x) : 12 cos (x) (sin (x) + 1)

12 cos (x) (sin (x) + 1) = 0

Solving each part separately

cos (x) = 0 or sin (x) + 1 = 0

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

sin (x) + 1 = 0, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}

- 5 = ((sin (x)) (cos (x)))

\frac{179}{sin ( x + 2 0 ^{\circ} )} = \frac{125}{sin ( 12 5 ^{\circ} )}

sin (x) = \frac{15}{12}

sin (z) = 5

cos (5 x) cos (x) + sin (5 x) sin (x) = 0

What is the general solution for 12cos(x)+6sin(2x)=0,0<= x<= 2pi ?
The general solution for 12cos(x)+6sin(2x)=0,0<= x<= 2pi is x= pi/2 ,x=(3pi)/2