Question

2cosx-3=2secx

Answer 1

What is the question?

Answer 2

find "x" I gather

Answer 3

Find all values of x to the nearest degree that satisfy the equation in the interval 0 ≤ x ≤ 360

Answer 4

\(\bf 2cos(x)-3=2sec(x)\qquad recall\qquad sec(\theta)=\cfrac{1}{cos(\theta)}\quad thus\\ \quad \\
2cos(x)-3=2sec(x)\implies 2cos(x)-3=2\cdot \cfrac{1}{cos(x)}\\ \quad \\2cos(x)-3=2sec(x)\implies 2cos(x)-3=\cfrac{2}{cos(x)}\\ \quad \\
cos(x)[2cos(x)-3]=2\implies 2cos^2(x)-3cos(x)-2=0\\ \quad \\
\begin{array}{llll}
\color{blue}{2ab^2}&\bf \color{blue}{-3b^2}&\bf \color{blue}{-2=0}\quad \textit{thus just solve the quadratic}\\
2cos^2(x)&-3cos(x)&-2=0
\end{array}\)

Answer 5

hmm

Answer 6

\(\bf \bf 2cos(x)-3=2sec(x)\qquad recall\qquad sec(\theta)=\cfrac{1}{cos(\theta)}\quad thus\\ \quad \\
2cos(x)-3=2sec(x)\implies 2cos(x)-3=2\cdot \cfrac{1}{cos(x)}\\ \quad \\2cos(x)-3=2sec(x)\implies 2cos(x)-3=\cfrac{2}{cos(x)}\\ \quad \\
cos(x)[2cos(x)-3]=2\implies 2cos^2(x)-3cos(x)-2=0\\ \quad \\
\begin{array}{llll}
\color{blue}{2ab^2}&\bf \color{blue}{-3b}&\bf \color{blue}{-2=0}\quad \textit{thus just solve the quadratic}\\
2cos^2(x)&-3cos(x)&-2=0
\end{array}\)

Answer 7

once you get what cos(x) is, then you can just get the inverse cosine to both sides to get the angle "x"

Answer 8

Thank you!