What is the general solution for 3cos(A)=sin^2(A) ?

The general solution for 3cos(A)=sin^2(A) is A=1.26319…+2pin,A=2pi-1.26319…+2pin

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3cos(A)=sin^2(A)

3 cos (A) = sin^{2} (A)

A = 1.26319 \dots + 2 πn, A = 2 π - 1.26319 \dots + 2 πn

Solution steps

3 cos (A) = sin^{2} (A)

Subtract

sin^{2} (A)

from both sides

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

Rewrite using trig identities

- 1 + cos^{2} (A) + 3 cos (A) = 0

Solve by substitution

cos (A) = \frac{- 3 + 13}{2}, cos (A) = \frac{- 3 - 13}{2}

cos (A) = \frac{- 3 + 13}{2} : A = arccos (\frac{- 3 + 13}{2}) + 2 πn, A = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

cos (A) = \frac{- 3 - 13}{2} :

No Solution

Combine all the solutions

A = arccos (\frac{- 3 + 13}{2}) + 2 πn, A = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

Show solutions in decimal form

A = 1.26319 \dots + 2 πn, A = 2 π - 1.26319 \dots + 2 πn

\frac{sin ( x )}{tan ( x )} = - 0.375

sin (2 x) = - sin (x), 0 \leq x \leq 2 π

2 sin (x) cos (x) + 2 sin (x) + cos (x) + 1 = 0

sin^{2} (θ) = 0

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

What is the general solution for 3cos(A)=sin^2(A) ?
The general solution for 3cos(A)=sin^2(A) is A=1.26319…+2pin,A=2pi-1.26319…+2pin