What is the general solution for 1+sin(2a)=sin^2(a) ?

The general solution for 1+sin(2a)=sin^2(a) is a= pi/2+2pin,a=(3pi)/2+2pin,a=-0.46364…+pin

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1+sin(2a)=sin^2(a)

1 + sin (2 a) = sin^{2} (a)

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn, a = - 0.46364 \dots + πn

Solution steps

1 + sin (2 a) = sin^{2} (a)

Subtract

sin^{2} (a)

from both sides

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

Rewrite using trig identities

cos^{2} (a) + 2 cos (a) sin (a) = 0

Factor

cos^{2} (a) + 2 cos (a) sin (a) : cos (a) (cos (a) + 2 sin (a))

cos (a) (cos (a) + 2 sin (a)) = 0

Solving each part separately

cos (a) = 0 or cos (a) + 2 sin (a) = 0

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

cos (a) + 2 sin (a) = 0 : a = arctan (- \frac{1}{2}) + πn

Combine all the solutions

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn, a = arctan (- \frac{1}{2}) + πn

Show solutions in decimal form

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn, a = - 0.46364 \dots + πn

\frac{( cos ^{3} ( a ) )}{( 2 cos ^{2} ( a ) - 1 )} = cos (a)

cos (x - 4 5^{\circ}) = 0

7 tan^{2} (x) - 15 = 0

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

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

What is the general solution for 1+sin(2a)=sin^2(a) ?
The general solution for 1+sin(2a)=sin^2(a) is a= pi/2+2pin,a=(3pi)/2+2pin,a=-0.46364…+pin