Question

Hello, can you please help me solve question six !

Answer 1

QUESTION A

The equation is given to be:

$\cos x-\sec x=1$

Step 1: Apply the identity:

$\sec x=\frac{1}{\cos x}$

Therefore, we have:

$\cos x-\frac{1}{\cos x}=1$

Step 2: Multiply all through by cos x:

$(\cos x)^2-1=\cos x$

Step 3: Rewrite the equation

$(\cos x)^2-\cos x-1=0$

Step 4: Make the substitution for cos x = m

$\begin{gathered} \therefore \\ m^2-m-1=0 \end{gathered}$

Step 5: Solve the quadratic equation using the quadratic formula

$\begin{gathered} m=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=1,b=-1,c=-1 \\ m=\frac{-(-1)\pm\sqrt[]{(-1)^2-4\times1\times(-1)}}{2\times1} \\ m=\frac{1\pm\sqrt[]{1+4}}{2} \\ m=\frac{1\pm\sqrt[]{5}}{2} \\ \therefore \\ m=\frac{1+\sqrt[]{5}}{2},\frac{1-\sqrt[]{5}}{2} \end{gathered}$

Step 6: Substitute for m back into the solution of the quadratic equation

$\begin{gathered} \cos x=\frac{1+\sqrt[]{5}}{2} \\ or \\ \cos x=\frac{1-\sqrt[]{5}}{2} \end{gathered}$

Step 7: Solve for x by finding the cosine inverse of the solutions

$x=\cos ^{-1}(\frac{1+\sqrt[]{5}}{2})=\text{ undefined}$

or

$x=\cos ^{-1}(\frac{1-\sqrt[]{5}}{2})=128.17$

The value of x is 128.17.

QUESTION B

The equation is:

$\cos x+\sec x=1$

Step 1: Rewrite the equation

$\cos x+\frac{1}{\cos x}=1$

Step 2: Multiply all through by cos x

$(\cos x)^2+1=\cos x$

Step 3: Rearrange the equation terms

$(\cos x)^2-\cos x+1=0$

Step 4: Make the substitution for cos x = m

$m^2-m+1=0$

Step 5: Solve the quadratic equation

$m=\frac{1+i\sqrt[]{3}}{2},\frac{1-i\sqrt[]{3}}{2}$

Step 6: Substitute for m back into the solution of the quadratic equation

$\begin{gathered} \cos x=\frac{1+i\sqrt[]{3}}{2} \\ or \\ \cos x=\frac{1-i\sqrt[]{3}}{2} \end{gathered}$

Step 7: Solve for x by finding the cosine inverse of the solutions

$\begin{gathered} x=\cos ^{-1}(\frac{1+i\sqrt[]{3}}{2})=\text{ undefined} \\ or \\ c=\cos ^{-1}(\frac{1-i\sqrt[]{3}}{2})=\text{ undefined} \end{gathered}$

There is no solution for x.