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

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

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cos^2(2x)+cos(2x)-3/4 =0

cos^{2} (2 x) + cos (2 x) - \frac{3}{4} = 0

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

Solution steps

cos^{2} (2 x) + cos (2 x) - \frac{3}{4} = 0

Solve by substitution

cos (2 x) = \frac{1}{2}, cos (2 x) = - \frac{3}{2}

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

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

No Solution

Combine all the solutions

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

sin^{2} (\frac{x}{2}) = a

cos (θ) = \frac{11}{5}

10 cos (θ) = 4 sin^{2} (θ) + 2

\frac{tan ^{2} ( x ) ( 1 + cot ^{2} ( x ) )}{sec ( x )} = cos (x)

4 cos^{2} (x) tan (x) - tan (x) = 0

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