Question

А)3cos(5x )=√2; б)−2 cos( x\6 )=√3.

Answer 1

$3cos(5x)=\sqrt{2} \\cos(5x)=\frac{\sqrt{2} }{3} \\\\5x=\pm arccos(\frac{\sqrt{2} }{3})+2\pi k , k \in Z \ (:5)\\x=\frac{\pm arccos(\frac{\sqrt{2} }{3}) }{5}+\frac{2\pi k }{5}, k \in Z$

$-2cos(\frac{x}{6})=\sqrt{3} \\cos(\frac{x}{6} )=-\frac{\sqrt{3} }{2} \\\frac{x}{6}=\pm(\pi -\frac{\pi }{6})+2\pi k , k \in Z \ (*6)\\x=\pm 5\pi +12\pi k , k \in Z$

Answer 2

To solve the given equations, we'll isolate the cosine terms and then solve for the value of 'x'. Here's how to do it step by step:

а) 3cos(5x) = √2

Step 1: Isolate the cosine term. Divide both sides by 3: cos(5x) = √2 / 3

Step 2: Solve for 'x'. To solve for 'x', take the inverse cosine (arccos) of both sides: x = arccos(√2 / 3)

Note: The inverse cosine function gives us the principal value between 0 and π (180 degrees). However, this equation may have multiple solutions depending on the interval of 'x' required.

б) -2cos(x/6) = √3

Step 1: Isolate the cosine term. Divide both sides by -2: cos(x/6) = -√3 / 2

Step 2: Solve for 'x'. To solve for 'x', take the inverse cosine (arccos) of both sides: x/6 = arccos(-√3 / 2)

Step 3: Solve for 'x'. To find the value of 'x', multiply both sides by 6: x = 6 * arccos(-√3 / 2)

As before, please note that the inverse cosine function gives us the principal value between 0 and π (180 degrees), but this equation may have multiple solutions depending on the interval of 'x' required.