we have the equation
[tex]-9\cos x=12\sec x-21[/tex]
Rewrite
[tex]-9\cos x=\frac{12}{\cos x}-21[/tex]
Multiply both sides by cosx
[tex]-9\cos x\cdot\cos x=\cos x\cdot\frac{12}{\cos x}-21\cdot\cos x[/tex][tex]\begin{gathered} -9\cos ^2x=12-21\cos x \\ -9\cos ^2x+21\cos x-12=0 \end{gathered}[/tex]
Change the variable
u=cosx
substitute
[tex]-9u^2+21u-12=0[/tex]
Solve the quadratic equation
using the formula
a=-9
b=21
c=-12
substitute
[tex]u=\frac{-21\pm\sqrt[]{21^2-4(-9)(-12)}}{2(-9)}[/tex][tex]\begin{gathered} u=\frac{-21\pm\sqrt[]{9}}{-18} \\ \\ u=\frac{-21\pm3}{-18} \end{gathered}[/tex]
The values of u are
u=1 and u=4/3
Remember that
u=cosx
For u=1
the interval is [0,360) ----> the value of 360 degrees is not included
cosx=1 ------> the value of x=0 degrees
For u=4/3
cosx=4/3 -------> is not a solution (cosine cannot be greater than 1)
therefore
The solution is x=0 degrees
