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

The general solution for (1+sin(x))/(cos^2(x))=2 is x= pi/6+2pin,x=(5pi)/6+2pin

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

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

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

Solution steps

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

Subtract

2

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = - 1

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

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

5 sin^{2} (θ) + 2 sin (θ) = 0

cos (θ) cos (3 θ) - sin (θ) sin (3 θ) = 0

1 - 2 sin (θ) = cos (2 θ), 0 \leq θ \leq 2 π

2 sin^{2} (θ) + 5 sin (θ) - 12 = 0

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

What is the general solution for (1+sin(x))/(cos^2(x))=2 ?
The general solution for (1+sin(x))/(cos^2(x))=2 is x= pi/6+2pin,x=(5pi)/6+2pin