What is the general solution for 9tan^2(z)cos(z)=3cos(z) ?

The general solution for 9tan^2(z)cos(z)=3cos(z) is z=(5pi)/6+pin,z= pi/6+pin

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9tan^2(z)cos(z)=3cos(z)

9 tan^{2} (z) cos (z) = 3 cos (z)

z = \frac{5 π}{6} + πn, z = \frac{π}{6} + πn

Solution steps

9 tan^{2} (z) cos (z) = 3 cos (z)

Subtract

3 cos (z)

from both sides

9 tan^{2} (z) cos (z) - 3 cos (z) = 0

Factor

9 tan^{2} (z) cos (z) - 3 cos (z) : 3 cos (z) (3 tan (z) + 1) (3 tan (z) - 1)

3 cos (z) (3 tan (z) + 1) (3 tan (z) - 1) = 0

Solving each part separately

cos (z) = 0 or 3 tan (z) + 1 = 0 or 3 tan (z) - 1 = 0

cos (z) = 0 : z = \frac{π}{2} + 2 πn, z = \frac{3 π}{2} + 2 πn

3 tan (z) + 1 = 0 : z = \frac{5 π}{6} + πn

3 tan (z) - 1 = 0 : z = \frac{π}{6} + πn

Combine all the solutions

z = \frac{π}{2} + 2 πn, z = \frac{3 π}{2} + 2 πn, z = \frac{5 π}{6} + πn, z = \frac{π}{6} + πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

z = \frac{5 π}{6} + πn, z = \frac{π}{6} + πn

sin (2 x) sin (4 x) = cos (2 x) cos (4 x)

\frac{1}{sec ( θ )} = - \frac{7}{25}

3 tan^{3} (x) - 2 tan^{2} (x) - 32 tan (x) - 32 = 0

x, 2 sin (5 x) - 1 = 0

cos (\frac{x}{2}) = \frac{1}{2}

What is the general solution for 9tan^2(z)cos(z)=3cos(z) ?
The general solution for 9tan^2(z)cos(z)=3cos(z) is z=(5pi)/6+pin,z= pi/6+pin