Mister Exam
sin(2x+pi/4)≥sin3pi/4 inequation
The solution
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/ pi\ sin(3*pi) sin|2*x + --| >= --------- \ 4 / 4
$$\sin{\left(2 x + \frac{\pi}{4} \right)} \geq \frac{\sin{\left(3 \pi \right)}}{4}$$
sin(2*x + pi/4) >= sin(3*pi)/4
Detail solution
Given the inequality:
$$\sin{\left(2 x + \frac{\pi}{4} \right)} \geq \frac{\sin{\left(3 \pi \right)}}{4}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(2 x + \frac{\pi}{4} \right)} = \frac{\sin{\left(3 \pi \right)}}{4}$$
Solve:
Given the equation
$$\sin{\left(2 x + \frac{\pi}{4} \right)} = \frac{\sin{\left(3 \pi \right)}}{4}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x + \frac{\pi}{4} = 2 \pi n + \operatorname{asin}{\left(0 \right)}$$
$$2 x + \frac{\pi}{4} = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi$$
Or
$$2 x + \frac{\pi}{4} = 2 \pi n$$
$$2 x + \frac{\pi}{4} = 2 \pi n + \pi$$
, where n - is a integer
Move
$$\frac{\pi}{4}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = 2 \pi n - \frac{\pi}{4}$$
$$2 x = 2 \pi n + \frac{3 \pi}{4}$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n - \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
$$x_{1} = \pi n - \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{8}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{8} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(2 x + \frac{\pi}{4} \right)} \geq \frac{\sin{\left(3 \pi \right)}}{4}$$
$$\sin{\left(2 \left(\pi n - \frac{\pi}{8} - \frac{1}{10}\right) + \frac{\pi}{4} \right)} \geq \frac{\sin{\left(3 \pi \right)}}{4}$$
but
Then
$$x \leq \pi n - \frac{\pi}{8}$$
no execute
one of the solutions of our inequality is:
$$x \geq \pi n - \frac{\pi}{8} \wedge x \leq \pi n + \frac{3 \pi}{8}$$
$$\sin{\left(2 x + \frac{\pi}{4} \right)} \geq \frac{\sin{\left(3 \pi \right)}}{4}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(2 x + \frac{\pi}{4} \right)} = \frac{\sin{\left(3 \pi \right)}}{4}$$
Solve:
Given the equation
$$\sin{\left(2 x + \frac{\pi}{4} \right)} = \frac{\sin{\left(3 \pi \right)}}{4}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x + \frac{\pi}{4} = 2 \pi n + \operatorname{asin}{\left(0 \right)}$$
$$2 x + \frac{\pi}{4} = 2 \pi n - \operatorname{asin}{\left(0 \right)} + \pi$$
Or
$$2 x + \frac{\pi}{4} = 2 \pi n$$
$$2 x + \frac{\pi}{4} = 2 \pi n + \pi$$
, where n - is a integer
Move
$$\frac{\pi}{4}$$
to right part of the equation
with the opposite sign, in total:
$$2 x = 2 \pi n - \frac{\pi}{4}$$
$$2 x = 2 \pi n + \frac{3 \pi}{4}$$
Divide both parts of the equation by
$$2$$
$$x_{1} = \pi n - \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
$$x_{1} = \pi n - \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
This roots
$$x_{1} = \pi n - \frac{\pi}{8}$$
$$x_{2} = \pi n + \frac{3 \pi}{8}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n - \frac{\pi}{8}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{\pi}{8} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(2 x + \frac{\pi}{4} \right)} \geq \frac{\sin{\left(3 \pi \right)}}{4}$$
$$\sin{\left(2 \left(\pi n - \frac{\pi}{8} - \frac{1}{10}\right) + \frac{\pi}{4} \right)} \geq \frac{\sin{\left(3 \pi \right)}}{4}$$
sin(-1/5 + 2*pi*n) >= 0
but
sin(-1/5 + 2*pi*n) < 0
Then
$$x \leq \pi n - \frac{\pi}{8}$$
no execute
one of the solutions of our inequality is:
$$x \geq \pi n - \frac{\pi}{8} \wedge x \leq \pi n + \frac{3 \pi}{8}$$
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Solving inequality on a graph
Rapid solution
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/ / / ___________\\ / / ___________\ \\ | | | / ___ || | | / ___ | || | | |\/ 2 + \/ 2 || | |\/ 2 - \/ 2 | || Or|And|0 <= x, x <= atan|--------------||, And|x <= pi, pi - atan|--------------| <= x|| | | | ___________|| | | ___________| || | | | / ___ || | | / ___ | || \ \ \\/ 2 - \/ 2 // \ \\/ 2 + \/ 2 / //
$$\left(0 \leq x \wedge x \leq \operatorname{atan}{\left(\frac{\sqrt{\sqrt{2} + 2}}{\sqrt{2 - \sqrt{2}}} \right)}\right) \vee \left(x \leq \pi \wedge \pi - \operatorname{atan}{\left(\frac{\sqrt{2 - \sqrt{2}}}{\sqrt{\sqrt{2} + 2}} \right)} \leq x\right)$$
((0 <= x)∧(x <= atan(sqrt(2 + sqrt(2))/sqrt(2 - sqrt(2)))))∨((x <= pi)∧(pi - atan(sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2))) <= x))
Rapid solution 2
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/ ___________\ / ___________\
| / ___ | | / ___ |
|\/ 2 + \/ 2 | |\/ 2 - \/ 2 |
[0, atan|--------------|] U [pi - atan|--------------|, pi]
| ___________| | ___________|
| / ___ | | / ___ |
\\/ 2 - \/ 2 / \\/ 2 + \/ 2 /
$$x\ in\ \left[0, \operatorname{atan}{\left(\frac{\sqrt{\sqrt{2} + 2}}{\sqrt{2 - \sqrt{2}}} \right)}\right] \cup \left[\pi - \operatorname{atan}{\left(\frac{\sqrt{2 - \sqrt{2}}}{\sqrt{\sqrt{2} + 2}} \right)}, \pi\right]$$
x in Union(Interval(0, atan(sqrt(sqrt(2) + 2)/sqrt(2 - sqrt(2)))), Interval(pi - atan(sqrt(2 - sqrt(2))/sqrt(sqrt(2) + 2)), pi))