Question

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

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Answer 1

Step-by-step explanation:

Given:

$2 \sin(3x) + 3 \cos(3x) = 0$

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

$2 \sin(3x) = - 3 \cos(3x)$

Step 2: Cross multiply

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

Step 3: Write equation in terms of tan 3x

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

Step 4: Take inverse tan both sides

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

Step 5: Cross multiply 3 to RHS

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

Step 6: General solution can be written as

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

Step 7:General solution in radian form

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

Final answer:

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

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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