Question

Find the number of solutions of |cos xl = sin x, 0<X<4r​

Answer 1

Required Method

• Modulus Equation

The value of a modulus function is always non-negative and we have one on the left-hand side. It refers to the distance from the origin, on the number line.

• Property of Modulus Functions

For real values of $x$, the modulus satisfies $|x|^{2}=x^{2}$, and this can be used to solve our trigonometric equation. Beware of extraneous solutions. $^{[1]}$

Correct Question

Find the number of solutions of equation ① in $0<x<4\pi$.

Equation ① $|\cos x|=\sin x$

Solution

➊ Solving the equation.

① $|\cos x|=\sin x$

To solve the equation, let's use the property of the modulus function. Squaring both sides we obtain the following trigonometric equation.

$\implies \cos^{2}x=\sin^{2} x$

$\implies 1-\sin^{2}x=\sin^{2} x$

$\implies 2\sin^{2} x=1$

$\therefore \sin^{2} x=\dfrac{1}{2}$

➋ Finding the solutions.

Now let's find the number of the solutions.

$\sin x=\dfrac{\sqrt{2}}{2}$ is a required solution, as $\sin x=|\cos x|\geq 0$.

$\sin x=-\dfrac{\sqrt{2} }{2}$ is an extraneous solution.

This implies the following.

$\implies x=\dfrac{\pi}{4} +2\pi n\ \text{or}\ x=\dfrac{3\pi}{4}+2\pi n$ where $n$ is an integer.

➌ Finding the number of solutions.

Where $n=0$ or $n=1$, there are two solutions respectively. So, the number of solutions is 4.

Final Result

There are four solutions in $0<x<4\pi$. $^{[2]}$

More Information

$^{[1]}$ The extraneous solutions arise from squaring both sides, since $A=B$ implies $A^{2}=B^{2}$ but not always $A^{2}=B^{2}$ implies $A=B$. It implies $A=\pm B$. We are solving the second case, so extraneous solutions can arise.

$^{[2]}$ Verification using the graph of $y=|\cos x|$ and $y=\sin x$ is in the attachment.