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

The general solution for cos(x)tan^2(x)=cos(x) is x=(3pi)/4+pin,x= pi/4+pin

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cos(x)tan^2(x)=cos(x)

cos (x) tan^{2} (x) = cos (x)

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Solution steps

cos (x) tan^{2} (x) = cos (x)

Subtract

cos (x)

from both sides

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

Factor

tan^{2} (x) cos (x) - cos (x) : cos (x) (tan (x) + 1) (tan (x) - 1)

cos (x) (tan (x) + 1) (tan (x) - 1) = 0

Solving each part separately

cos (x) = 0 or tan (x) + 1 = 0 or tan (x) - 1 = 0

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

tan (x) + 1 = 0 : x = \frac{3 π}{4} + πn

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Since the equation is undefined for:

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

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

1 + 2 sin (\frac{x ^{2}}{2}) = 0

2 sin^{2} (x) + 7 cos (x) = 5

2 sin^{2} (θ) - 3 sin (θ) + 1 = 0, 0 \leq θ < 2 π

2 sin^{2} (x) = 3 sin (x)

4 sin^{2} (x) - 3 = 0, 0 \leq x \leq 2 π

What is the general solution for cos(x)tan^2(x)=cos(x) ?
The general solution for cos(x)tan^2(x)=cos(x) is x=(3pi)/4+pin,x= pi/4+pin