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

The general solution for tan^2(x)-sin(x)=tan^2(x)sin^2(x) is x=2pin,x=pi+2pin

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tan^2(x)-sin(x)=tan^2(x)sin^2(x)

tan^{2} (x) - sin (x) = tan^{2} (x) sin^{2} (x)

x = 2 πn, x = π + 2 πn

Solution steps

tan^{2} (x) - sin (x) = tan^{2} (x) sin^{2} (x)

Subtract

tan^{2} (x) sin^{2} (x)

from both sides

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

Express with sin, cos

\frac{sin ^{2} ( x ) - sin ^{4} ( x ) - cos ^{2} ( x ) sin ( x )}{cos ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin^{2} (x) - sin^{4} (x) - cos^{2} (x) sin (x) = 0

Factor

sin^{2} (x) - sin^{4} (x) - cos^{2} (x) sin (x) : sin (x) (sin (x) - sin^{3} (x) - cos^{2} (x))

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

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

Combine all the solutions

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

Since the equation is undefined for:

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

x = 2 πn, x = π + 2 πn

3 cos (a) - 1 = 0

tan^{2} (x) + tan (x) + cot (x) + cot^{2} (x) = 4

tan (a) = 0

cos^{2} (x) + 3 ∣ cos (x) ∣ - 1 = 0

cos^{5} (x) = sin (7 5^{\circ})

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