12. 0/1 points details my notes larpcalclim4hs 5.3.038. find all solutions of the equation in the interval…

Explanation:

Step1: Rewrite in terms of sine and cosine

Recall that $\sec x=\frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$. The equation $2\sec x+2\tan x = 2$ becomes $\frac{2}{\cos x}+\frac{2\sin x}{\cos x}=2$. Multiply through by $\cos x$ (assuming $\cos x\neq0$) to get $2 + 2\sin x=2\cos x$. Then, $1+\sin x=\cos x$.

Step2: Square both sides

$(1 + \sin x)^2=\cos^{2}x$. Expand the left - hand side: $1 + 2\sin x+\sin^{2}x=\cos^{2}x$. Since $\cos^{2}x=1-\sin^{2}x$, we have $1 + 2\sin x+\sin^{2}x=1-\sin^{2}x$.

Step3: Rearrange the equation

Move all terms to one side: $1 + 2\sin x+\sin^{2}x-1+\sin^{2}x = 0$. Combine like terms: $2\sin^{2}x+2\sin x=0$. Factor out $2\sin x$: $2\sin x(\sin x + 1)=0$.

Step4: Solve for $\sin x$

Set each factor equal to zero. If $2\sin x=0$, then $\sin x = 0$. In the interval $[0,2\pi)$, $x = 0,\pi$. If $\sin x+1 = 0$, then $\sin x=-1$. In the interval $[0,2\pi)$, $x=\frac{3\pi}{2}$.

Step5: Check for extraneous solutions

Substitute $x = 0$ into the original equation: $2\sec(0)+2\tan(0)=2\times1 + 2\times0=2$, which is a valid solution. Substitute $x=\pi$ into the original equation: $2\sec(\pi)+2\tan(\pi)=2\times(-1)+2\times0=-2\neq2$, so $x = \pi$ is an extraneous solution. Substitute $x=\frac{3\pi}{2}$ into the original equation: $\sec x$ and $\tan x$ are undefined at $x=\frac{3\pi}{2}$, so $x=\frac{3\pi}{2}$ is an extraneous solution.

Answer:

$0$