What is the general solution for (cos^2(x)+1)/(1+cot^2(x))=1 ?

The general solution for (cos^2(x)+1)/(1+cot^2(x))=1 is x= pi/2+2pin,x=(3pi)/2+2pin

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(cos^2(x)+1)/(1+cot^2(x))=1

\frac{cos ^{2} ( x ) + 1}{1 + cot ^{2} ( x )} = 1

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

Solution steps

\frac{cos ^{2} ( x ) + 1}{1 + cot ^{2} ( x )} = 1

Subtract

1

from both sides

\frac{cos ^{2} ( x ) + 1}{1 + cot ^{2} ( x )} - 1 = 0

Simplify

\frac{cos ^{2} ( x ) + 1}{1 + cot ^{2} ( x )} - 1 : \frac{cos ^{2} ( x ) - cot ^{2} ( x )}{1 + cot ^{2} ( x )}

\frac{cos ^{2} ( x ) - cot ^{2} ( x )}{1 + cot ^{2} ( x )} = 0

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

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

Factor

cos^{2} (x) - cot^{2} (x) : (cos (x) + cot (x)) (cos (x) - cot (x))

(cos (x) + cot (x)) (cos (x) - cot (x)) = 0

Solving each part separately

cos (x) + cot (x) = 0 or cos (x) - cot (x) = 0

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

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

Combine all the solutions

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

3 sin^{2} (c) - 7 sin (x) + 2 = 0

1 + cos^{2} (x) = sin^{4} (x)

cos (t) = cos (2 t)

arctan (x + 2) = arcsin (\frac{7}{25}) + arccos (\frac{4}{5})

sec (a) = \frac{38}{13}

What is the general solution for (cos^2(x)+1)/(1+cot^2(x))=1 ?
The general solution for (cos^2(x)+1)/(1+cot^2(x))=1 is x= pi/2+2pin,x=(3pi)/2+2pin