To solve the trigonometric equation 6sin(x) + 5cos(x) - 2 = 0, we can rewrite it in terms of a single trigonometric function.
First, note that sin^2(x) + cos^2(x) = 1 (identity of trigonometric functions).
Now, let's multiply both sides of the equation by √(6^2 + 5^2) = √61 to get rid of the coefficients in front of sin(x) and cos(x) terms:
√61 6sin(x) + √61 5cos(x) - √61 * 2 = 06√61sin(x) + 5√61cos(x) - 2√61 = 0
Now, we can rewrite the left side of the equation as 6√61[sin(x)cos(α) + cos(x)sin(α)], where cos(α) = 6/√61 and sin(α) = 5/√61:
6√61[sin(x)cos(α) + cos(x)sin(α)] - 2√61 = 06√61sin(x+α) - 2√61 = 0
Therefore, the solution to the equation 6sin(x) + 5cos(x) - 2 = 0 is sin(x + α) = 2/3√61, where α is the angle whose sin and cos values are determined by the coefficients in front of sin(x) and cos(x) terms.
To solve the trigonometric equation 6sin(x) + 5cos(x) - 2 = 0, we can rewrite it in terms of a single trigonometric function.
First, note that sin^2(x) + cos^2(x) = 1 (identity of trigonometric functions).
Now, let's multiply both sides of the equation by √(6^2 + 5^2) = √61 to get rid of the coefficients in front of sin(x) and cos(x) terms:
√61 6sin(x) + √61 5cos(x) - √61 * 2 = 0
6√61sin(x) + 5√61cos(x) - 2√61 = 0
Now, we can rewrite the left side of the equation as 6√61[sin(x)cos(α) + cos(x)sin(α)], where cos(α) = 6/√61 and sin(α) = 5/√61:
6√61[sin(x)cos(α) + cos(x)sin(α)] - 2√61 = 0
6√61sin(x+α) - 2√61 = 0
Therefore, the solution to the equation 6sin(x) + 5cos(x) - 2 = 0 is sin(x + α) = 2/3√61, where α is the angle whose sin and cos values are determined by the coefficients in front of sin(x) and cos(x) terms.