Question

Solve the following equation:
4sin^2 (x) +7cos(x) = 6

Answer 1

Answer on Question #51409 – Math – Trigonometry

$$4 \sin^ {2} (x) + 7 \cos (x) = 6$$

Solution

Pythagorean trigonometric identity

$$\sin^ {2} (x) + \cos^ {2} (x) = 1$$

gives

$$\sin^ {2} (x) = 1 - \cos^ {2} (x).$$

Substitute for the initial equation

$$4 \sin^ {2} (x) + 7 \cos (x) = 6$$

and obtain

$$4 (1 - \cos^ {2} (x)) + 7 \cos (x) = 6, \text{ open brackets } 4 - 4 \cos^ {2} (x) + 7 \cos (x) = 6;$$

collect similar terms

$$4 \cos^ {2} (x) - 7 \cos (x) + 2 = 0.$$

Make a substitution

$$\cos (x) = t$$

such that $-1\leq t\leq 1$

After substitution, equation becomes as follows

$$4 t ^ {2} - 7 t + 2 = 0.$$

To solve it, calculate

$$\begin{array}{l} D = 7 ^ {2} - 4 \cdot 4 \cdot 2 = 49 - 32 = 17; \\ \left\{ \begin{array}{c} t _ {1} = \frac {7 - \sqrt {17}}{2 \cdot 4} = \frac {7 - \sqrt {17}}{8} < \frac {7 - \sqrt {16}}{8} = \frac {7 - 4}{8} = \frac {3}{8} < 1 \\ t _ {2} = \frac {7 + \sqrt {17}}{2 \cdot 4} = \frac {7 + \sqrt {17}}{8} > \frac {7 + \sqrt {16}}{8} = \frac {7 + 4}{8} = \frac {11}{8} > \frac {8}{8} = 1 \end{array} \right. \\ \end{array}$$

Because $t_2 = \frac{7 + \sqrt{17}}{8} > 1$ does not satisfy the condition $-1 \leq t \leq 1$ , skip this value, equation $\cos(x) = t_2$ has no roots.

Because $t_1 = \frac{7 - \sqrt{17}}{8} < 1$ satisfies the condition $-1 \leq t \leq 1$ , equation $\cos(x) = t_1$ has roots.

We have

$$\cos (x) = \frac {7 - \sqrt {1 7}}{8} \Rightarrow x = \cos^ {- 1} \left(\frac {7 - \sqrt {1 7}}{8}\right) + 2 \pi n,$$

where $\cos^{-1}(x)$ is the inverse cosine function, $n$ is integer.

Answer: $x = 2\pi n + \cos^{-1}\left(\frac{7 - \sqrt{17}}{8}\right), n \in \mathbb{Z}$ .

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