14cos2𝑥 + sin 2𝑥 = 6.
To solve the equation \( 14 \cos(2x) + \sin(2x) = 6 \), we will first rewrite it in terms of a single trigonometric function or use a substitution.
Rearranging the Equation: Move all terms to one side: \[ 14 \cos(2x) + \sin(2x) - 6 = 0 \]
Using a substitution: Let \( y = \cos(2x) \). Then, \( \sin(2x) = \sqrt{1 - y^2} \) (using the Pythagorean identity).
Substitute this into the equation: \[ 14y + \sqrt{1 - y^2} - 6 = 0 \] Rearranging gives: \[ \sqrt{1 - y^2} = 6 - 14y \] Now, square both sides to eliminate the square root (note that squaring both sides can introduce extraneous solutions, so we need to check our solutions at the end): \[ 1 - y^2 = (6 - 14y)^2 \] Expanding the right side: \[ 1 - y^2 = 36 - 168y + 196y^2 \] Moving all terms to one side: \[ 1 - 36 + 168y - 196y^2 - y^2 = 0 \] Combining like terms: \[ -197y^2 + 168y - 35 = 0 \] Rearranging gives: \[ 197y^2 - 168y + 35 = 0 \]
Using the Quadratic Formula: Now we will solve this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 197 \), \( b = -168 \), and \( c = 35 \): \[ y = \frac{168 \pm \sqrt{(-168)^2 - 4 \cdot 197 \cdot 35}}{2 \cdot 197} \] Calculating the discriminant: \[ (-168)^2 = 28224 \] \[ 4 \cdot 197 \cdot 35 = 27640 \] \[ 28224 - 27640 = 584 \] Continuing with the solution: \[ y = \frac{168 \pm \sqrt{584}}{394} \] Simplifying \(\sqrt{584}\): \[ \sqrt{584} = \sqrt{4 \cdot 146} = 2\sqrt{146} \] Thus: \[ y = \frac{168 \pm 2\sqrt{146}}{394} = \frac{84 \pm \sqrt{146}}{197} \]
Determining the values of \(y\): We have two possible \(y\) values: \[ y_1 = \frac{84 + \sqrt{146}}{197} \] \[ y_2 = \frac{84 - \sqrt{146}}{197} \]
Checking if the values satisfy \(-1 \leq y \leq 1\): Calculate \(y_1\) and \(y_2\) approximately: \[ \sqrt{146} \approx 12.083 \] \[ y_1 \approx \frac{84 + 12.083}{197} \approx \frac{96.083}{197} \approx 0.488 \] \[ y_2 \approx \frac{84 - 12.083}{197} \approx \frac{71.917}{197} \approx 0.364 \]
Both \(y_1\) and \(y_2\) are between -1 and 1, so they are valid solutions for \( \cos(2x) \).
- Finding \(x\): Solve for \(2x\): \[ 2x = \cos^{-1}(y_1) \quad \text{and} \quad 2x = \cos^{-1}(y_2) \] So: \[ x = \frac{1}{2} \cos^{-1}(y_1) + k\pi \quad \text{and} \quad x = \frac{1}{2} \cos^{-1}(y_2) + k\pi \] where \(k\) is any integer.
Make sure to check all conditions and values as necessary!
