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Question Number 150058 by mathdanisur last updated on 09/Aug/21

$$\mathrm{Solve}\mathrm{the}\mathrm{equation}:$$$${\cos }^4\left(\mathrm{x}\right)+\mathrm{i}{\sin }^4\left(\mathrm{x}\right)={4\mathrm{e}}^{4\mathbf{ix}}$$

Commented by MJS_new last updated on 10/Aug/21

$$\mathrm{I}\mathrm{found}\mathrm{these}(n\in \mathbb{Z}):$$$$x\approx n\pi +.866946376+.432060893\mathrm{i}$$$$x\approx n\pi +.00142216354+.301698277\mathrm{i}$$$$x\approx n\pi -.768959527+.828028492\mathrm{i}$$$$x\approx n\pi -1.26957042+.336555887\mathrm{i}$$

Commented by mathdanisur last updated on 10/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

Answered by MJS_new last updated on 10/Aug/21

$$\frac{\sqrt{2}}{4}⩽\mid {\cos }^{4}x+\mathrm{i}{\sin }^{4}x\mid ⩽1$$$$\mid {4\mathrm{e}}^{4\mathrm{i}x}\mid =4$$$$\Rightarrow \mathrm{no}\mathrm{solution}\in \mathbb{R}$$

Commented by mathdanisur last updated on 09/Aug/21

$$\mathrm{Thankyou}\mathbf{S}\mathrm{er}$$$$\mathrm{Dear}\mathbf{S}\mathrm{er},\mathrm{no}\mathrm{solution}\mathrm{in}\mathrm{real}\mathrm{numbers},\mathrm{but}\mathrm{it}$$$$\mathrm{has}\mathrm{a}\mathrm{complex}\mathrm{number}\mathrm{solution}$$

Answered by MJS_new last updated on 10/Aug/21

$${\cos }^{4}x+\mathrm{i}{\sin }^{4}x={4\mathrm{e}}^{4\mathrm{i}x}$$$${\cos }^{4}x+\mathrm{i}{\sin }^{4}x=4\cos 4x+4\mathrm{i}\sin 4x$$$$\mathrm{let}t=\tan x$$$$\frac{1}{{(t^2+1)}^2}+\frac{t^4}{{(t^2+1)}^2}\mathrm{i}=\frac{4(t^2-2t-1)(t^2+2t-1)}{{(t^2+1)}^2}-\frac{16t(t-1)(t+1)}{{(t^2+1)}^2}\mathrm{i}$$$$\mathrm{this}\mathrm{leads}\mathrm{to}$$$$t^4+\frac{16(1-4\mathrm{i})}{17}{t}^3-\frac{24(4+\mathrm{i})}{17}{t}^2-\frac{16(1-4\mathrm{i})}{17}t+\frac{3(4+\mathrm{i})}{17}=0$$$$\mathrm{I}\mathrm{found}\mathrm{no}\mathrm{useful}\mathrm{exact}\mathrm{solution}$$$$t_1\approx .799102359+.790210727\mathrm{i}$$$$t_2\approx .00130018452+.292866540\mathrm{i}$$$$t_3\approx -.363754201+.918561793\mathrm{i}$$$$t_4\approx -1.37782481+1.76306682\mathrm{i}$$$$t=\tan x\iff x=n\pi +\arctan t$$$$\Rightarrow \mathrm{the}\mathrm{solutions}\mathrm{I}\mathrm{posted}\mathrm{above}$$

Commented by mathdanisur last updated on 10/Aug/21

$$\mathrm{Thank}\mathrm{You}\mathbf{S}\mathrm{er}$$

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