If \( \sin x=0.6 \), evaluate \( 2 \cos x+3 \sin x \) , find \( \cos \theta \)

We wish to write the expression   2 cos x + 3 sin x in the form   R cos(x – θ), where R > 0 and θ is an angle (usually taken in the first quadrant). Step 1. Find R. Since a cosine–sine combination can be written as   A cos x + B sin x = R cos(x – θ), we have   R = √(A² + B²). Here A = 2 and B = 3, so   R = √(2² + 3²) = √(4 + 9) = √13. Step 2. Determine cos θ and sin θ. When rewriting the expression as R cos(x – θ), one standard way is to set   cos θ = A/R  and  sin θ = B/R. Thus,   cos θ = 2/√13  and  sin θ = 3/√13. One may also note that tan θ = (sinθ)/(cosθ) = 3/2. Step 3. Rewrite the expression. Therefore, the given expression becomes   2 cos x + 3 sin x = √13 cos(x – θ), with θ satisfying cos θ = 2/√13 and sin θ = 3/√13. Step 4. Use the given sin x = 0.6. Now, if we are also told that sin x = 0.6, then—assuming x lies in an interval where cosine is positive (typically the first quadrant)—we have   cos x = √(1 – sin² x) = √(1 – 0.36) = √0.64 = 0.8. Evaluating the expression directly gives   2 cos x + 3 sin x = 2(0.8) + 3(0.6)           = 1.6 + 1.8           = 3.4. Thus, when sin x = 0.6 (and cos x = 0.8), the value of 2 cos x + 3 sin x is 3.4. Summary of Answers: • When sin x = 0.6 (with cos x = 0.8), 2 cos x + 3 sin x = 3.4. • The angle shift θ is given by cos θ = 2/√13 (and sin θ = 3/√13).