What is the general solution for cos^2(x+30)= 1/4 ?

The general solution for cos^2(x+30)= 1/4 is x=360n+30,x=360n+270,x=360n+90,x=360n+210

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cos^2(x+30)= 1/4

cos^{2} (x + 3 0^{\circ}) = \frac{1}{4}

x = 36 0^{\circ} n + 3 0^{\circ}, x = 36 0^{\circ} n + 27 0^{\circ}, x = 36 0^{\circ} n + 9 0^{\circ}, x = 36 0^{\circ} n + 21 0^{\circ}

Solution steps

cos^{2} (x + 3 0^{\circ}) = \frac{1}{4}

Solve by substitution

cos (x + 3 0^{\circ}) = \frac{1}{2}, cos (x + 3 0^{\circ}) = - \frac{1}{2}

cos (x + 3 0^{\circ}) = \frac{1}{2} : x = 36 0^{\circ} n + 3 0^{\circ}, x = 36 0^{\circ} n + 27 0^{\circ}

cos (x + 3 0^{\circ}) = - \frac{1}{2} : x = 36 0^{\circ} n + 9 0^{\circ}, x = 36 0^{\circ} n + 21 0^{\circ}

Combine all the solutions

x = 36 0^{\circ} n + 3 0^{\circ}, x = 36 0^{\circ} n + 27 0^{\circ}, x = 36 0^{\circ} n + 9 0^{\circ}, x = 36 0^{\circ} n + 21 0^{\circ}

cos (5 7^{\circ}) = sin (x)

3 cos (x) = 2 - cos (x)

2 cos^{2} (3 x) - 3 cos (3 x) = - 1, 0 \leq x \leq 2 π

x, z = y^{s i n (x)}

t, 10 = 14 + 8 sin (\frac{π t}{12})

What is the general solution for cos^2(x+30)= 1/4 ?
The general solution for cos^2(x+30)= 1/4 is x=360n+30,x=360n+270,x=360n+90,x=360n+210