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

The general solution for 2cos^2(4x)+1=0 is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

Solve by substitution

cos (4 x) = i \frac{1}{2}, cos (4 x) = - i \frac{1}{2}

cos (4 x) = i \frac{1}{2} :

No Solution

cos (4 x) = - i \frac{1}{2} :

No Solution

Combine all the solutions

No Solution for x \in R

1 + cos (2 θ) = - 3 sin (θ)

2 cos (θ) + sin (2 θ) = 0

tan (x) = \frac{6 \cdot 3}{6}

cos^{2} (x) + sin^{2} (x) - 2 sin (x) = 0

sin (x) = cos^{2} (x) - sin^{2} (x) - 1

What is the general solution for 2cos^2(4x)+1=0 ?
The general solution for 2cos^2(4x)+1=0 is No Solution for x\in\mathbb{R}