Funções Trigonométricas

por BatataLaranja345 Sáb 13 Mar 2021, 16:12

Boa tarde amigos e amigas do fórum! Gostaria de saber quem poderia me ajudar com essa questão! Segue:

Calcule o valor de (secx)^⅔ +(tgx)^⅔, sabendo que:

√[sec²x + ∛(sec⁴x .tg²x)] + √[tg²x + ∛(sec²x . tg⁴x)] = 8

a) 1
b) 2
c) 3
d) 4
e) 5

Quem puder me ajudar, agradeço desde já! Abraços! cheers

por **Renan Almeida** Ter 16 Mar 2021, 13:58

[latex]\sec^{2/3}{x} + \tan^{2/3}{x}[/latex]

[latex]\frac{1 + \sin^{2/3}{x}}{\cos^{2/3}{x}}[/latex]

[latex]\sqrt{\sec^{2}{x}+\sqrt[3]{\sec^{4}{x}\tan^{2}{x}}} + \sqrt{\tan^{2}{x}+\sqrt[3]{\sec^{2}{x}\tan^{4}{x}}} = 8[/latex]

[latex]\sqrt{\frac{1}{\cos^{2}{x}}+\sqrt[3]{\frac{1}{\cos^{4}{x}}\frac{\sin^{2}{x}}{\cos^{2}{x}}}} + \sqrt{\frac{\sin^{2}{x}}{\cos^{2}{x}}+\sqrt[3]{\frac{1}{\cos^{2}{x}}\frac{\sin^{4}{x}}{\cos^{4}{x}}}} = 8[/latex]

[latex]\sqrt{\frac{1}{\cos^{2}{x}}+\sqrt[3]{\frac{\sin^{2}{x}}{\cos^{6}{x}}}} + \sqrt{\frac{\sin^{2}{x}}{\cos^{2}{x}}+\sqrt[3]{\frac{\sin^{4}{x}}{\cos^{6}{x}}}} = 8[/latex]

[latex]\sqrt{\frac{1}{\cos^{2}{x}}+\frac{\sin^{2/3}{x}}{cos^{2}{x}}} + \sqrt{\frac{\sin^{2}{x}}{\cos^{2}{x}}+\frac{\sin^{4/3}{x}}{cos^{2}{x}}} = 8[/latex]

[latex]\frac{\sqrt{1 + sin^{2/3}{x}}}{\cos{x}} + \frac{\sqrt{\sin^{4/3}{x}(1 + \sin^{2/3}{x})}}{\cos{x}} = 8[/latex]

[latex]\frac{\sqrt{1 + sin^{2/3}{x}}}{\cos{x}} + \frac{\sin^{2/3}{x}\sqrt{1 + \sin^{2/3}{x}}}{\cos{x}} = 8[/latex]

[latex]\frac{\sqrt{1 + sin^{2/3}{x}}(1+\sin^{2/3}{x})}{\cos{x}} = 8[/latex]

[latex]\sqrt{1 + \sin^{2/3}{x}} = y[/latex]

[latex]y*y^2 = 8\cos{x}[/latex]

[latex]y = \sqrt[3]{8\cos{x}} = 2 \sqrt[3]{\cos{x}}[/latex]

[latex]y^2 = 4\cos^{2/3}{x}[/latex]

[latex]\frac{1 + \sin^{2/3}{x}}{\cos^{2/3}{x}}[/latex]

[latex]\frac{y^2}{\cos^{2/3}{x}}[/latex]

[latex]\frac{4\cos^{2/3}{x}}{\cos^{2/3}{x}}[/latex]

[latex]4[/latex] (D)