Question

Q1.Find all the values of x, in between the range [0,2π] for the following equation.
2sin(3x)+3cos(3x)=0.

Answer 1

Step-by-step explanation:

Given:

$\begin{gathered}2 \sin(3x) + 3 \cos(3x) = 0 \\\end{gathered}$

To find:Find all the values of x, in between the range [0,2π] for the equation.

Solution:

Step 1: Take 3 cos(3x) to other side

$\begin{gathered}2 \sin(3x) = - 3 \cos(3x) \\\end{gathered}$

Step 2: Cross multiply

$\begin{gathered}\frac{ \sin(3x) }{ \cos(3x) } = \frac{ - 3}{2} \\\end{gathered}$

Step 3: Write equation in terms of tan 3x

$\begin{gathered}\tan(3x) = \frac{ - 3}{2} \\\end{gathered}$

Step 4: Take inverse tan both sides

$\begin{gathered}3x = {tan}^{ - 1} ( \frac{ - 3}{2} ) \\\end{gathered}$

Step 5: Cross multiply 3 to RHS

$\begin{gathered}x = \frac{1}{3} {tan}^{ - 1} ( \frac{ - 3}{2} ) \\\end{gathered}$

Step 6: General solution can be written as

$\begin{gathered}x = -\frac{1}{3} {tan}^{ - 1} ( \frac{3}{2} )+\frac{n \pi }{3} \\\end{gathered}$

Step 7:General solution in radian form

$\begin{gathered}x = -\frac{0.98279}{3}+\frac{n \pi }{3} \\\end{gathered}$

Final answer:

$\begin{gathered}\bold{\red{x = -\frac{0.98279}{3}+\frac{n \pi }{3}}} \\\end{gathered}$

One can put different values of n for exact solution.

Hope it helps you.

Note*: It is clear that angle is negative,so it is not lies between [0,2π].

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1) Verify Cauchy's mean value theorem for the function sin x and cos x in the interval [a, b]

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Answer 2

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