Answer to Question #273152 in Trigonometry for seiji

Question #273152

Find the solutions of the equation if 0≀𝑑<2πœ‹

tan^2 π‘‘βˆ’1=0



1
Expert's answer
2021-11-30T04:53:42-0500
tan⁑2tβˆ’1=0\tan^2t-1=0

(tan⁑tβˆ’1)(tan⁑t+1)=0(\tan t-1)(\tan t+1)=0

tan⁑t=±1\tan t=\pm1

t=Ο€4+Ο€n2,n∈Zt=\dfrac{\pi}{4}+\dfrac{\pi n}{2}, n\in \Z

If we consider 0≀t<2Ο€0\leq t<2\pi

t1=Ο€4,t2=3Ο€4,t3=5Ο€4,t4=7Ο€4t_1=\dfrac{\pi}{4}, t_2=\dfrac{3\pi}{4}, t_3=\dfrac{5\pi}{4}, t_4=\dfrac{7\pi}{4}

{Ο€4,3Ο€4,5Ο€4,7Ο€4}\{\dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}\}


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