What is the general solution for cos(2θ)=2,cos^2(θ)-1=2,(-3/5)^2-1=-7/25 ?

The general solution for cos(2θ)=2,cos^2(θ)-1=2,(-3/5)^2-1=-7/25 is No Solution for θ\in\mathbb{R}

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cos(2θ)=2,cos^2(θ)-1=2,(-3/5)^2-1=-7/25

cos (2 θ) = 2, cos^{2} (θ) - 1 = 2, (- \frac{3}{5})^{2} - 1 = - \frac{7}{25}

No Solution for θ \in R

Solution steps

cos (2 θ) = 2, cos^{2} (θ) - 1 = 2, (- \frac{3}{5})^{2} - 1 = - \frac{7}{25}

- 1 \leq cos (x) \leq 1

No Solution for θ \in R

2 sin (2 θ) = 1, 0 \leq θ \leq 2 π

2 cos^{2} (x) - 3 sin (x) = 3

14 cos (θ) - 5 = 5 cos (θ) - 5

- 4 sin (2 θ) = 5 cos (2 θ)

tan (A) = \frac{3}{4}

What is the general solution for cos(2θ)=2,cos^2(θ)-1=2,(-3/5)^2-1=-7/25 ?
The general solution for cos(2θ)=2,cos^2(θ)-1=2,(-3/5)^2-1=-7/25 is No Solution for θ\in\mathbb{R}