Problem 48 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following equations and tick the correct one. If \(0<\theta<2 \pi\) and \(2 \sin ^{2} \theta-5 \sin \theta+2>0\), then the range of \(\theta\) is (a) \(\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)\) (b) \(\left(0, \frac{5 \pi}{6}\right) \cup(\pi, 2 \pi)\) (c) \(\left(0, \frac{\pi}{6}\right) \cup(\pi, 2 \pi)\) (d) None.

Step-by-Step Solution

Answer

The solution of the inequality \(2 \sin ^{2} \theta-5 \sin \theta+2>0\) for \(0<\theta<2 \pi\) is \((0, \pi/2)\), which do not match any of the given options. So, the answer will be (d) None.

Step 1: Write the inequality in standard form

The inequality is already given in the standard quadratic form \(ax^{2}+bx+c > 0\), where \(x = \sin \theta\), \(a = 2\), \(b = -5\), and \(c = 2\). Now we'll solve \(2 \sin ^{2} \theta-5 \sin \theta+2 = 0\) to find the solutions, which act as turning points for the inequality.

Step 2: Solving the quadratic equation

The equation formed is \(2 \sin ^{2} \theta -5 \sin \theta + 2 = 0\). This is of the form \(ax^{2} + bx + c = 0\), where \(x = \sin \theta \). Now we solve for \(\sin \theta\) using the quadratic formula \(x = [-b\pm \sqrt{b^{2}-4ac}]/2a\). Plugging the values of \(a = 2\), \(b = -5\), and \(c = 2\), the solution for \(\sin \theta\) comes out to be \(1\) and \(2\). However, since the value of \(\sin \theta\) lies between -1 and 1, the root \(2\) isn't possible. So, \(\sin \theta = 1\).

Step 3: Finding theta

To find the turning points \(\theta\), we need to solve \(\sin \theta = 1\). The value of \(\theta\) is \(\frac{\pi}{2}\). Since \(0 < \theta < 2 \pi\), the only solution obtained is \(\theta = \frac{\pi}{2}\).

Step 4: Drawing the intervals

Draw all the points on the number line and divide it into intervals. The points are \(0\), \(\pi/2\), and \(2\pi\). This divides the interval \(0<\theta<2\pi\) into two sub-intervals: \((0, \pi/2)\), \((\pi/2, 2\pi)\). Now, select a test point in each interval and substitute in the inequality to check whether the inequality is satisfied.

Step 5: Testing the intervals

Take the test points \(\frac{\pi}{4}\) for the interval \((0, \pi/2)\) and \(\frac{3\pi}{2}\) for the interval \((\pi/2, 2\pi)\) for \(\theta\). Substituting \(\theta = \frac{\pi}{4}\) in the inequality, it holds true and for \(\theta = \frac{3\pi}{2}\), it doesn't hold true. Thus the solution of given inequality is \((0, \pi/2)\).

Step 6: Check the options

None of the given options include the interval \((0, \pi/2)\) entirely within them. Therefore, the correct answer is (d) None.

Key Concepts

Solving Trigonometric Equations

Understanding how to solve trigonometric equations is critical in dealing with inequalities in trigonometry. A trigonometric equation is one that involves trigonometric functions such as sine, cosine, or tangent. The goal is to find the angles (or values of \( \theta \) in this case) that make the equation true within the given interval, which is often between \( 0 \) and \( 2\pi \) in the context of standard trigonometric problems.

To solve trigonometric equations, one often uses algebraic techniques such as factoring and applying the quadratic formula, as well as trigonometric identities like the Pythagorean identity. When the equation resembles a quadratic form, such as \( 2 \sin ^{2} \theta-5 \sin \theta+2 = 0 \), we can treat \( \sin \theta \) as a variable and solve as we would a standard quadratic equation. This involves finding possible values of \( \sin \theta \) that satisfy the equation and then determining the corresponding angles \( \theta \) within the specified interval.

Quadratic Inequalities

Quadratic inequalities are solved using methods similar to solving quadratic equations but with extra consideration for the inequality sign. Such inequalities take the form \( ax^{2} + bx + c \gt 0 \) or \( ax^{2} + bx + c \lt 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) is the variable.

To solve a quadratic inequality, start by finding the roots of the corresponding quadratic equation \( ax^{2} + bx + c = 0 \) as these roots divide the number line into intervals. Then, one can select test points from each interval and substitute them back into the inequality to check which intervals satisfy the inequality. The solutions of the inequality are the intervals for which the tests yield true results. In the context of trigonometric inequalities, we use this method to determine the ranges of the trigonometric function that satisfy the inequality.

Inequalities in Trigonometry

Solving inequalities in trigonometry involves determining the range of angles that satisfy the given trigonometric inequality. It's important to remember that the solutions to trigonometric inequalities are typically intervals or unions of intervals rather than discrete angles.

After solving the corresponding trigonometric equation, as done in earlier steps to find the critical points, you must consider the periodic nature of trigonometric functions. Draw a number line or a unit circle to illustrate the intervals between these points. This visual aid allows you to test values within the intervals to see if they satisfy the original inequality. The beauty of trigonometric functions is their cyclic patterns, which help in predicting the intervals that will satisfy the inequality. Always keep in mind the restrictions on trigonometric values, such as \( -1 \leq \sin \theta \leq 1 \), when considering possible solutions.