Problem 60 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

\(\cos ^{2} x+\sin x+1=0\)

Step-by-Step Solution

Answer

The equation \(\cos^2 x + \sin x +1= 0\) has one solution \(x = \frac{3\pi}{2} + 2k\pi\), where k is an integer.

Step 1: Substitute \(\cos^2 x\) with its equivalent form.

With the formula \(\cos^2 x = 1 - \sin^2 x\), the given equation becomes \(1 - \sin^2 x + \sin x + 1 = 0\). Summing up like terms, the equation can then be written as \(-\sin^2 x + \sin x + 2 = 0\).

Step 2: Assign variable

For easier handling, let a new variable \(y = \sin x\). So, the above equation can now be written in terms of \(y\): \(-y^2 + y + 2 = 0\).

Step 3: Convert to standard form

Our equation now reads as a quadratic equation. To convert it into the standard quadratic equation format \(ax^2 + bx + c = 0\), we multiply the equation by -1. This gives us: \(y^2 - y - 2 = 0\).

Step 4: Solve the quadratic equation

The quadratic equation can be factored to \(y^2 -2y + y - 2 = 0\), which can further be grouped and factored to \(y(y - 2) + 1(y - 2) = 0\) and then \((y-2)(y+1) = 0\). Setting each factor equal to zero gives the solutions \(y = 2, -1\).

Step 5: Substitute \(y\) with \(\sin x\)

Since \(y = \sin x\), we substitute \(y\) with \(sin x\). Giving us \(\sin x = 2, -1\). However, a sine value cannot be greater than 1 or less than -1. So, the equation \(\sin x = 2\) has no solution while \(\sin x = -1\) leads to \(x = \frac{3\pi}{2} + 2k\pi\), where \(k\) is an integer.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. These equations often take the standard form:

• \[ ax^2 + bx + c = 0 \]

Here, \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. The solutions to quadratic equations can be found using several methods, such as:

• Factoring
• The quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• Completing the square

In the solution of the trigonometric equation given in the exercise, the original trigonometric equation is transformed into a quadratic form by substituting \(\sin x = y\). This helps simplify the equation into a familiar format, which is easier to solve.
By using substitution, we erected a bridge from the trigonometric world to the easier-to-solve algebraic realm of quadratic equations.

Sine Function

The sine function, denoted as \(\sin x\), is one of the most useful functions in trigonometry, describing the relationship between an angle in a right-angled triangle and the ratio of the length of the side opposite the angle to the hypotenuse. It is periodic with a period of \(2\pi\) and oscillates between -1 and 1:

• The general behavior: \(\sin x\) has a smooth, wave-like graph.
• Its range is from -1 to 1, meaning it never exceeds these values.
• Its period is \(2\pi\), repeating every \(360^{\circ}\).

In the exercise, the relevance of the sine function is clear since \(\sin x\) determines the possibility of finding a solution. The transformation from \(\cos^2 x\) to \(1 - \sin^2 x\) leverages the Pythagorean identity, making it possible to express everything in terms of \(\sin x\). Thus, when solving for \(\sin x = 2\) and \(\sin x = -1\), only the latter provides a valid solution because \(\sin x = 2\) is outside the allowed range.

Factorization

Factorization is a critical algebraic skill that involves breaking down an expression into a product of simpler expressions, often called factors. When dealing with quadratic equations, factorization can be a straightforward way to find roots if the equation is factorable.
In the exercise, the quadratic equation generated, \(y^2 - y - 2 = 0\), is factored by first rewriting it with terms like \(-2y + y\) and identifying common factors. The factorization process involves:

• Group terms: \(y^2 - 2y + y - 2 = 0\)
• Identify common factors and group: \(y(y - 2) + 1(y - 2) = 0\)
• Factor completely: \((y - 2)(y + 1) = 0\)

Setting each factor equal to zero, \((y - 2) = 0\) and \((y + 1) = 0\), yields possible solutions for \(y\). Applying these to the trigonometric context, we evaluate accepted values based on the range of the sine function, leading us to valid solutions while discarding any extraneous ones.