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)

Solution

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

Solution

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

Degrees

a = 9 0^{\circ} + 36 0^{\circ} n, a = 27 0^{\circ} + 36 0^{\circ} n, a = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

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

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

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

1 + sin (2 a)

Use the Pythagorean identity:

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

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

Use the Double Angle identity:

sin (2 a) = 2 sin (a) cos (a)

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

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

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

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

Simplify

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

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

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = sin (a), b = cos (a)

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

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

Add similar elements:

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

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

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

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

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

cos^{2} (a) = cos (a) cos (a)

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

Factor out common term

cos (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) = 0

General solutions for

cos (a) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

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

Rewrite using trig identities

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

Divide both sides by

cos (a), cos (a) \neq = 0

\frac{cos ( a ) + 2 sin ( a )}{cos ( a )} = \frac{0}{cos ( a )}

Simplify

1 + \frac{2 sin ( a )}{cos ( a )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 + 2 tan (a) = 0

1 + 2 tan (a) = 0

Move

1

to the right side

1 + 2 tan (a) = 0

Subtract

1

from both sides

1 + 2 tan (a) - 1 = 0 - 1

Simplify

2 tan (a) = - 1

2 tan (a) = - 1

Divide both sides by

2

2 tan (a) = - 1

Divide both sides by

2

\frac{2 tan ( a )}{2} = \frac{- 1}{2}

Simplify

tan (a) = - \frac{1}{2}

tan (a) = - \frac{1}{2}

Apply trig inverse properties

tan (a) = - \frac{1}{2}

General solutions for

tan (a) = - \frac{1}{2}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

a = arctan (- \frac{1}{2}) + πn

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

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Frequently Asked Questions (FAQ)

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