the function $g$ is defined by $g(x)=sec^{2}x+\tan x$. what are all solutions to $g(x) = 1$ on the interval…

the function $g$ is defined by $g(x)=sec^{2}x+\tan x$. what are all solutions to $g(x) = 1$ on the interval $0leq xleq2pi$?\n(a) $x = 0,x=\frac{3pi}{4},x=pi,x=\frac{7pi}{4}$, and $x = 2pi$ only\n(b) $x=\frac{pi}{4},x=\frac{pi}{2},x=\frac{5pi}{4}$, and $x=\frac{3pi}{2}$ only\n(c) $x=pi k$ and $x =-\frac{pi}{4}+pi k$, where $k$ is any integer\n(d) $x=\frac{pi}{2}+pi k$ and $x=\frac{pi}{4}+pi k$, where $k$ is any integer

the function $g$ is defined by $g(x)=sec^{2}x+\tan x$. what are all solutions to $g(x) = 1$ on the interval $0leq xleq2pi$?\n(a) $x = 0,x=\frac{3pi}{4},x=pi,x=\frac{7pi}{4}$, and $x = 2pi$ only\n(b) $x=\frac{pi}{4},x=\frac{pi}{2},x=\frac{5pi}{4}$, and $x=\frac{3pi}{2}$ only\n(c) $x=pi k$ and $x =-\frac{pi}{4}+pi k$, where $k$ is any integer\n(d) $x=\frac{pi}{2}+pi k$ and $x=\frac{pi}{4}+pi k$, where $k$ is any integer

Answer

Answer:

A. $x = 0,x=\frac{3\pi}{4},x = \pi,x=\frac{7\pi}{4}$, and $x = 2\pi$ only

Explanation:

Step1: Use trigonometric identity

Recall $\sec^{2}x=1 + \tan^{2}x$. So $g(x)=1+\tan^{2}x+\tan x$.

Step2: Set up the equation

Set $g(x) = 1$, then $1+\tan^{2}x+\tan x=1$.

Step3: Simplify the equation

Subtract 1 from both sides: $\tan^{2}x+\tan x=0$.

Step4: Factor the equation

Factor out $\tan x$: $\tan x(\tan x + 1)=0$.

Step5: Solve for $\tan x$

Case 1: $\tan x=0$. In the interval $0\leq x\leq2\pi$, $x = 0,\pi,2\pi$. Case 2: $\tan x=- 1$. In the interval $0\leq x\leq2\pi$, $x=\frac{3\pi}{4},\frac{7\pi}{4}$.