Question #16170

solve, (sec^(2)θ - cos^(2)θ)^(2) = tanθ - sin^(2)θ , between the range 0<θ<360"

Expert's answer

(1cos2θcos2θ)2=tanθsin2θ\left(\frac {1}{\cos^ {2} \theta} - \cos^ {2} \theta\right) ^ {2} = \tan \theta - \sin^ {2} \theta


We can solve it graphically:



And we have next solutions:


{θ=πk,kZθ=2(πm1.31182),mZθ=2(πn+0.258974),nZθ={0,0.517948,π,3.65954,2π}.θ[0,2π]\left\{ \begin{array}{l} \theta = \pi k, k \in Z \\ \theta = 2 (\pi m - 1.31182), m \in Z \\ \theta = 2 (\pi n + 0.258974), n \in Z \Rightarrow \theta = \{0, 0.517948, \pi, 3.65954, 2\pi \}. \\ \theta \in [0, 2\pi] \end{array} \right.


Answer: θ={0,0.517948,π,3.65954,2π}\theta = \{0, 0.517948, \pi, 3.65954, 2\pi\} .

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