Problem 36
Question
Solve the following inequalities: $$ 4\left(\tan ^{-1} x\right)^{2}-8\left(\tan ^{-1} x\right)+3<0 $$
Step-by-Step Solution
Verified Answer
The solution to the inequality is \(\tan(0.5) < x < \tan(1.5)\).
1Step 1: Rewrite in the form of a typical quadratic inequality
Rewrite the inequality into a normal quadratic form. The inequality becomes \(4u^2 - 8u + 3 < 0\), where \(u = \tan^{-1}(x)\).
2Step 2: Find the roots of the quadratic
Factor the quadratic to find its roots. That gives \((2u - 1)(2u - 3) < 0\), so we have two solutions, \(u = 0.5\) and \(u = 1.5\).
3Step 3: Solve for \(\tan^{-1}(x)\)
Replace \(u\) with \(\tan^{-1}(x)\) to obtain \(\tan^{-1}(x) = 0.5\) and \(\tan^{-1}(x) = 1.5\). Solving these two equations will give \(x = \tan(0.5)\) and \(x = \tan(1.5)\).
4Step 4: Determine the intervals of x
The inequality \(4u^2 - 8u + 3 < 0\) is only true between the roots, i.e. \(0.5 < \tan^{-1}(x) < 1.5\). In terms of x, this means \(\tan(0.5) < x < \tan(1.5)\).
Key Concepts
Solving Trigonometric InequalitiesInverse Trigonometric FunctionsQuadratic Equation Roots
Solving Trigonometric Inequalities
Trigonometric inequalities, such as \(4(\tan^{-1} x)^2 - 8(\tan^{-1} x) + 3 < 0\), involve finding the values of variable where the inequality holds true. Here is how you approach such problems:
- Rewrite the trigonometric inequality in terms of a simpler variable to handle, like \(u = \tan^{-1}(x)\), transforming it into a regular algebraic quadratic inequality which can be easier to solve.
- Determine the roots of the quadratic equation, as these will define the critical points or boundaries of the solution set.
- Using inverse trigonometric functions, you can find the corresponding angles and therefore the values of \(x\) that satisfy the original inequality.
- Analyze the signs of the various intervals around the critical points to find out where the original inequality holds true.
- Finally, interpret the solution in the context of the original variable, in this case, express the solution in terms of \(x\) using the tangent function.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \(\tan^{-1}(x)\), or arctan(x), are used to find the angle whose tangent is \(x\). They are pivotal when working with trigonometric inequalities because they allow us to go from an angle to a ratio and vice versa. Here are key points about inverse trigonometric functions:
- They reverse the action of the original trigonometric functions (like sine, cosine, and tangent).
- Because trigonometric functions are periodic, the inverse functions typically return the principal value - the smallest positive angle that satisfies the equation.
- They have a specific range and domain, ensuring that each input has a unique output.
- Understanding the graphical interpretation of these functions can aid in visualizing solutions to trigonometric inequalities.
Quadratic Equation Roots
Finding the roots of a quadratic equation is an essential part of solving trigonometric inequalities. Roots of the equation \(ax^2+bx+c=0\) determine where the graph of the equation intersects the x-axis, indicating values for which the equation equals zero. Here's what to keep in mind about quadratic roots:
- The roots can be found using factoring, completing the square, or the quadratic formula.
- The roots play a pivotal role in determining the intervals of the solution of the inequalities. Any quadratic inequality will divide the number line into intervals based on these roots. The solution to the inequality lies within one or more of these intervals.
- The sign of the quadratic expression changes at each root, which helps to determine on which side of the root the inequality holds true.
- In context, after finding the roots of the quadratic equation, such as in our exercise \(2u - 1\) and \(2u - 3\), we use the inverse trigonometric functions to map these roots back onto the x-values that solve the original inequality.
Other exercises in this chapter
Problem 35
Solve the following inequalities: $$ \tan ^{2}\left(\sin ^{-1} x\right)>1 $$
View solution Problem 35
If \(\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi\), then the value of \(\frac{1}{y z}+\frac{1}{z x}+\frac{1}{x y}\) is (a) 0 (b) 1 (c) \(\frac{1}{x y z}\) (d) \(
View solution Problem 36
If \(x
View solution Problem 37
Solve the following inequalities: $$ 4 \cot ^{-1} x-\left(\cot ^{-1} x\right)^{2}-3 \geq 0 $$
View solution