What is the general solution for cos(x)tan^2(x)-3cos(x)=0 ?

The general solution for cos(x)tan^2(x)-3cos(x)=0 is x=(2pi)/3+pin,x= pi/3+pin

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cos(x)tan^2(x)-3cos(x)=0

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

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

Solution steps

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

Factor

cos (x) tan^{2} (x) - 3 cos (x) : cos (x) (tan (x) + 3) (tan (x) - 3)

cos (x) (tan (x) + 3) (tan (x) - 3) = 0

Solving each part separately

cos (x) = 0 or tan (x) + 3 = 0 or tan (x) - 3 = 0

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

tan (x) + 3 = 0 : x = \frac{2 π}{3} + πn

tan (x) - 3 = 0 : x = \frac{π}{3} + πn

Combine all the solutions

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

Since the equation is undefined for:

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

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

2 sin (x) + 3 cot (x) - 3 csc (x) = 0

cos^{2} (θ) - 3 cos (θ) + 2 = 0

cos (2 x - \frac{π}{6}) = - \frac{1}{2}, \frac{π}{2} \leq x < π

\frac{( sin ( x ) )}{( 4.2 )} = \frac{( sin ( 9 5 ^{\circ} ) )}{( 10.26 )}

- sin (x) = - cos (x)

What is the general solution for cos(x)tan^2(x)-3cos(x)=0 ?
The general solution for cos(x)tan^2(x)-3cos(x)=0 is x=(2pi)/3+pin,x= pi/3+pin