Solve the following equation. $\tan ^ { 4 } x - 1 = 0$
A) $x = \frac { \pi } { 2 } + n \pi$ ,where n is an integer
B) $x = n \pi \text { and } x = \frac { 3 \pi } { 2 } + 2 n \pi$ ,where n is an integer
C) $x = n \pi \text { and } x = \frac { 3 \pi } { 4 } + n \pi$ ,where n is an integer
D) $x = \frac { \pi } { 4 } + \frac { n \pi } { 2 }$ ,where n is an integer
E) $x = n \pi \text { and } x = \frac { \pi } { 2 } + n \pi$ ,where n is an integer

Question

Solve the following equation. $\tan ^ { 4 } x - 1 = 0$
A)​ $x = \frac { \pi } { 2 } + n \pi$ ,where n is an integer
B) $x = n \pi \text { and } x = \frac { 3 \pi } { 2 } + 2 n \pi$ ,where n is an integer
C)​ $x = n \pi \text { and } x = \frac { 3 \pi } { 4 } + n \pi$ ,where n is an integer
D) $x = \frac { \pi } { 4 } + \frac { n \pi } { 2 }$ ,where n is an integer
E)​ $x = n \pi \text { and } x = \frac { \pi } { 2 } + n \pi$ ,where n is an integer

Quizplus · Accepted Answer

The equation $\tan^4 x - 1 = 0$ simplifies to $\tan^4 x = 1$, which means $\tan x = \pm 1$. The tangent function equals 1 at $x = \frac{\pi}{4}$ and equals -1 at $x = \frac{3\pi}{4}$, both within the first cycle of the tangent function. Since the tangent function has a period of $\pi$, the general solutions are $x = \frac{\pi}{4} + n\frac{\pi}{2}$ and $x = \frac{3\pi}{4} + n\frac{\pi}{2}$, which can be combined into $x = \frac{\pi}{4} + \frac{n\pi}{2}$ where $n$ is an integer.

Solve the Following Equation tan⁡4x−1=0\tan ^ { 4 } x - 1 = 0tan4x−1=0

Solve the Following Equation $\tan ^ { 4 } x - 1 = 0$