By expressing sqrt3sin x+3cos x in the form A sin(x+alpha), solve 4 sqrt3sin x+3cos x=sqrt3, for 0

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By expressing  sqrt3sin x+3cos x in the form  A sin(x+alpha), solve 4  sqrt3sin x+3cos x=sqrt3, for  0 <= x <= 2pi.

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Use Trigonometric Identity: We want to express sqrt(3)sin(x) + 3cos(x) in the form A sin(x+alpha). To do this, we will use the identity sin(x+alpha) = sin(x)cos(alpha) + cos(x)sin(alpha).Match Coefficients: Let A sin(x+alpha) = A(sin(x)cos(alpha) + cos(x)sin(alpha)). We want to match the coefficients of sin(x) and cos(x) to those in sqrt(3)sin(x) + 3cos(x).Solve Equations for A: Matching coefficients, we get A cos(alpha) = sqrt(3) and A sin(alpha) = 3. We can solve these two equations to find A and alpha.Find A Using Pythagorean Identity: To find A, we use the Pythagorean identity: A^2 = (A cos(alpha))^2 + (A sin(alpha))^2. Substituting the values we have A^2 = (sqrt(3))^2 + 3^2 = 3 + 9 = 12.Find A and Alpha: Taking the square root of both sides, we get A = sqrt(12) = 2 sqrt(3).Express in Desired Form: To find alpha, we take the ratio of the two equations: tan(alpha) = (A sin(alpha)) / (A cos(alpha)) = 3 / sqrt(3) = sqrt(3).Isolate sin(x + alpha): Using the arctan function, we find alpha = arctan(sqrt(3)). Since we are looking for solutions in the interval [0, 2pi], we consider the principal value of arctan(sqrt(3)), which is pi/3.Find Solutions for sin(y): Now we have expressed sqrt(3)sin(x) + 3cos(x) as 2 sqrt(3) sin(x + pi/3). We can now solve the equation 2 sqrt(3) sin(x + pi/3) = sqrt(3).Substitute x + alpha for y: Divide both sides by 2 sqrt(3) to isolate sin(x + pi/3): sin(x + pi/3) = sqrt(3) / (2 sqrt(3)) = 1/2.Solve for x: We look for solutions to sin(x + pi/3) = 1/2. The solutions to sin(y) = 1/2 are y = pi/6 + 2kpi and y = 5pi/6 + 2kpi, where k is an integer.Check Solutions in Interval: Substitute x + pi/3 for y: x + pi/3 = pi/6 + 2kpi or x + pi/3 = 5pi/6 + 2kpi. We solve for x in each case.Final Solutions: For the first equation: x = pi/6 - pi/3 + 2kpi = -pi/6 + 2kpi. For the second equation: x = 5pi/6 - pi/3 + 2kpi = pi/2 + 2kpi.We need to find the values of x in the interval [0, 2pi]. For the first equation, x = -pi/6 is not in the interval, but x = -pi/6 + 2pi = 11pi/6 is. For the second equation, x = pi/2 is in the interval.The solutions to the equation sqrt(3)sin(x) + 3cos(x) = sqrt(3) in the interval [0, 2pi] are x = 11pi/6 and x = pi/2.

By expressing $\newline$ $\sqrt{3}\sin x+3\cos x$ in the form $A \sin(x+\alpha)$ , solve $\sqrt{3}\sin x+3\cos x=\sqrt{3}$ , for $\newline$ $0 \leq x \leq 2\pi$ .

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By expressing \newline3sin⁡x+3cos⁡x\sqrt{3}\sin x+3\cos x3​sinx+3cosx in the form Asin⁡(x+α)A \sin(x+\alpha)Asin(x+α), solve3sin⁡x+3cos⁡x=3\sqrt{3}\sin x+3\cos x=\sqrt{3}3​sinx+3cosx=3​, for \newline0≤x≤2π0 \leq x \leq 2\pi0≤x≤2π.

By expressing $\newline$ $\sqrt{3}\sin x+3\cos x$ in the form $A \sin(x+\alpha)$ , solve $\sqrt{3}\sin x+3\cos x=\sqrt{3}$ , for $\newline$ $0 \leq x \leq 2\pi$ .