Question

Use trigonometric identities, algebraic methods, and inverse trigonometric functions, as necessary, to solve the following trigonometric equation on the interval [0, 2x),Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution."5cos(x) + 7cos(x) = -2

Answer 1

We can see that the expression is a quadratic expression of the form:

`5\cos ^2(x)+7\cos (x)+2=0`

Now we can say:

`\cos (x)=\tau`

And use the quadratic equation:

`\begin{gathered} 5\tau^2+7\tau+2=0 \\ \tau_(1,2)=\frac{-7\pm\sqrt[]{7^2-4\cdot5\cdot2}}{2\cdot5}=\frac{-7\pm\sqrt[]{49-40}}{10}=(-7\pm3)/(10) \\ \tau_1=(-7+3)/(10)=-(4)/(10)=-(2)/(5) \\ \tau_2=(-7-3)/(10)=-(10)/(10)=-1 \end{gathered}`

The solutions are the values of x such:

`\begin{gathered} \cos (x)=-1 \\ \cos (x)=-(2)/(5) \end{gathered}`

We know that if x = π, cos(x) = -1. Thus, π is a solution.

The other solutions are:

`\cos (x)=-(2)/(5)`

And since cos has a period of 2π, the solutions are:

The two other solutions for [0, 2pi) are:

`x=\cos ^(-1)(-(2)/(5))\approx1.9823`

And:

`x=2\pi-\cos ^(-1)(-(2)/(5))\approx4.3008`

All the solutions in the interval are:

`x=1.9823,\pi,4.3008^{}`