Question

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

Answer 1

Let y^2 = x^2 - 3x

y^2 - y - 2 = 0

y^2 - 2y + y - 2 = 0

y(y - 2) + 1 (y - 2) = 0

(y+1)(y-2) = 0

y = -1 or y = 2

Then

By resubstitution -

x^2 - 3x = -1

x^2 - 3x + 1 = 0

By shreedharacharya formula -

x = {3 ± √(9–4)}/4

x = ( 3 ± √5 ) / 4

Also -

x^2 - 3x = 2

x^2 - 3x - 2 = 0

By shreedharacharya formula -

x = {3 ± √(9+16)}/4

x = ( 3 ± 5 ) / 4

x = 2 or x = -1/2

x = 2, -1/2, (3+√5)/4, (3-√5)/4

A2A. THANK-YOU!

Answer 2

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:

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

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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