Mister Exam
sin(x+pi/6)>=-1/2 inequation
The solution
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/ pi\ sin|x + --| >= -1/2 \ 6 /
$$\sin{\left(x + \frac{\pi}{6} \right)} \geq - \frac{1}{2}$$
sin(x + pi/6) >= -1/2
Detail solution
Given the inequality:
$$\sin{\left(x + \frac{\pi}{6} \right)} \geq - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x + \frac{\pi}{6} = 2 \pi n + \operatorname{asin}{\left(- \frac{1}{2} \right)}$$
$$x + \frac{\pi}{6} = 2 \pi n - \operatorname{asin}{\left(- \frac{1}{2} \right)} + \pi$$
Or
$$x + \frac{\pi}{6} = 2 \pi n - \frac{\pi}{6}$$
$$x + \frac{\pi}{6} = 2 \pi n + \frac{7 \pi}{6}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation
with the opposite sign, in total:
$$x = 2 \pi n - \frac{\pi}{3}$$
$$x = 2 \pi n + \pi$$
$$x_{1} = 2 \pi n - \frac{\pi}{3}$$
$$x_{2} = 2 \pi n + \pi$$
$$x_{1} = 2 \pi n - \frac{\pi}{3}$$
$$x_{2} = 2 \pi n + \pi$$
This roots
$$x_{1} = 2 \pi n - \frac{\pi}{3}$$
$$x_{2} = 2 \pi n + \pi$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n - \frac{\pi}{3}\right) + - \frac{1}{10}$$
=
$$2 \pi n - \frac{\pi}{3} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(x + \frac{\pi}{6} \right)} \geq - \frac{1}{2}$$
$$\sin{\left(\left(2 \pi n - \frac{\pi}{3} - \frac{1}{10}\right) + \frac{\pi}{6} \right)} \geq - \frac{1}{2}$$
but
Then
$$x \leq 2 \pi n - \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$x \geq 2 \pi n - \frac{\pi}{3} \wedge x \leq 2 \pi n + \pi$$
$$\sin{\left(x + \frac{\pi}{6} \right)} \geq - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(x + \frac{\pi}{6} \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x + \frac{\pi}{6} = 2 \pi n + \operatorname{asin}{\left(- \frac{1}{2} \right)}$$
$$x + \frac{\pi}{6} = 2 \pi n - \operatorname{asin}{\left(- \frac{1}{2} \right)} + \pi$$
Or
$$x + \frac{\pi}{6} = 2 \pi n - \frac{\pi}{6}$$
$$x + \frac{\pi}{6} = 2 \pi n + \frac{7 \pi}{6}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation
with the opposite sign, in total:
$$x = 2 \pi n - \frac{\pi}{3}$$
$$x = 2 \pi n + \pi$$
$$x_{1} = 2 \pi n - \frac{\pi}{3}$$
$$x_{2} = 2 \pi n + \pi$$
$$x_{1} = 2 \pi n - \frac{\pi}{3}$$
$$x_{2} = 2 \pi n + \pi$$
This roots
$$x_{1} = 2 \pi n - \frac{\pi}{3}$$
$$x_{2} = 2 \pi n + \pi$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n - \frac{\pi}{3}\right) + - \frac{1}{10}$$
=
$$2 \pi n - \frac{\pi}{3} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(x + \frac{\pi}{6} \right)} \geq - \frac{1}{2}$$
$$\sin{\left(\left(2 \pi n - \frac{\pi}{3} - \frac{1}{10}\right) + \frac{\pi}{6} \right)} \geq - \frac{1}{2}$$
/1 pi \
-sin|-- + -- - 2*pi*n| >= -1/2
\10 6 / but
/1 pi \
-sin|-- + -- - 2*pi*n| < -1/2
\10 6 / Then
$$x \leq 2 \pi n - \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$x \geq 2 \pi n - \frac{\pi}{3} \wedge x \leq 2 \pi n + \pi$$
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