What is the general solution for 2cos^2(4x+1)=0 ?

The general solution for 2cos^2(4x+1)=0 is x=-1/4+pi/8+(pin)/2 ,x=-1/4+(3pi)/8+(pin)/2

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2cos^2(4x+1)=0

2 cos^{2} (4 x + 1) = 0

x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}, x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

Solution steps

2 cos^{2} (4 x + 1) = 0

Divide both sides by

2

cos (4 x + 1) = 0

General solutions for

cos (4 x + 1) = 0

4 x + 1 = \frac{π}{2} + 2 πn, 4 x + 1 = \frac{3 π}{2} + 2 πn

Solve

4 x + 1 = \frac{π}{2} + 2 πn : x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}

Solve

4 x + 1 = \frac{3 π}{2} + 2 πn : x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}, x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

x, 2 sin (x) = - 1

t, f = a cos (\frac{t π}{6} - \frac{11 π}{12})

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7, 0 \leq θ < 36 0^{\circ}

\frac{cos ( x )}{sin ^{2} ( x )} = \frac{2}{3}

7 cos (b) - 13 = 0

What is the general solution for 2cos^2(4x+1)=0 ?
The general solution for 2cos^2(4x+1)=0 is x=-1/4+pi/8+(pin)/2 ,x=-1/4+(3pi)/8+(pin)/2