What is the general solution for 4cos^2(A)-3=0 ?

The general solution for 4cos^2(A)-3=0 is A= pi/6+2pin,A=(11pi)/6+2pin,A=(5pi)/6+2pin,A=(7pi)/6+2pin

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4cos^2(A)-3=0

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

A = \frac{π}{6} + 2 πn, A = \frac{11 π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn, A = \frac{7 π}{6} + 2 πn

Solution steps

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

Solve by substitution

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

cos (A) = \frac{3}{2} : A = \frac{π}{6} + 2 πn, A = \frac{11 π}{6} + 2 πn

cos (A) = - \frac{3}{2} : A = \frac{5 π}{6} + 2 πn, A = \frac{7 π}{6} + 2 πn

Combine all the solutions

A = \frac{π}{6} + 2 πn, A = \frac{11 π}{6} + 2 πn, A = \frac{5 π}{6} + 2 πn, A = \frac{7 π}{6} + 2 πn

\frac{sin ( 4 2 ^{\circ} )}{6.25} = \frac{sin ( x )}{9}

\frac{tan ( x ) + cot ( x )}{csc ( x )} = cos (x)

tan (θ) = - \frac{3}{7}

sin (2 x) = 0, - π \leq x \leq π

cos (2 x) - sin (2 x) = 1

What is the general solution for 4cos^2(A)-3=0 ?
The general solution for 4cos^2(A)-3=0 is A= pi/6+2pin,A=(11pi)/6+2pin,A=(5pi)/6+2pin,A=(7pi)/6+2pin