# Math Help - simpl inequation

1. ## simpl inequation

Solve in R :

2. hello
$
2\sin (\frac{\pi}{6})=1
$

3. 2 sin( 2x - pi/6 ) - 1 <= 0
2 sin( 2x - pi/6 ) <= 1
sin( 2x - pi/6 ) <= 1/2
2x - pi/6 <= sin^-1(1/2)
2x = sin^-1(1/2) + pi/6
x = ( sin^-1(1/2) + pi/6 ) / 2

Therefore x = 15.26...

4. Originally Posted by justmaths
2 sin( 2x - pi/6 ) - 1 <= 0
2 sin( 2x - pi/6 ) <= 1
sin( 2x - pi/6 ) <= 1/2
2x - pi/6 <= sin^-1(1/2)
2x = sin^-1(1/2) + pi/6
x = ( sin^-1(1/2) + pi/6 ) / 2

Therefore x = 15.26...
There are several errors in this solution.

5. Originally Posted by Raoh
hello
$
2\sin (\frac{\pi}{6})=1
$
@OP: Expanding a little on this hint:

You can therefore write $\sin \left(2x - \frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right)$.

Now solve for $x$ using the fact that if sin A = sin B then:

Case 1: $A = B + 2n \pi$

Case 2: $A = (\pi - B) + 2n \pi$

where n is an integer.

Now that you know the values of x for which $2 \sin \left(2x - \frac{\pi}{6}\right) - 1 = 0$ you should be able to write down the intervals over which $2 \sin \left(2x - \frac{\pi}{6}\right) - 1 \leq 0$.

(drawing a graph might help ....)

6. Originally Posted by dhiab
Solve in R :
An alternative approach:

First solve $2 \sin \left(2x - \frac{\pi}{6}\right) - 1 = 0$:

$\sin \left(2x - \frac{\pi}{6}\right) = \frac{1}{2}$

Case 1: $2x - \frac{\pi}{6} = \frac{\pi}{6} + 2n \pi$

Case 2: $2x - \frac{\pi}{6} = \frac{5\pi}{6} + 2n \pi$

where n is an integer.

Now solve for x and then solve the inequality as explained in my previous post.

7. Originally Posted by mr fantastic
An alternative approach:

First solve $2 \sin \left(2x - \frac{\pi}{6}\right) - 1 = 0$:

$\sin \left(2x - \frac{\pi}{6}\right) = \frac{1}{2}$

Case 1: $2x - \frac{\pi}{6} = \frac{\pi}{6} + 2n \pi$

Case 2: $2x - \frac{\pi}{6} = \frac{5\pi}{6} + 2n \pi$

where n is an integer.

Now solve for x and then solve the inequality as explained in my previous post.
Hello : You can solve the inequality by trigo-circle Thank you

8. Originally Posted by dhiab
Hello: you have inequation et not equation Thank you
Yes, we understood that! Apparently you did not understand the solution. Since the functions here are continuous, they can only change from "<" to ">" and vice-versa at "=". First solve the equation, then, as mr. fantastic said, check the intervals between the solutions to the equation to see where it is ">" and "<".