Question

Solve ${\sin ^2}x - {\cos ^2}x = 0$ for $x$ in the interval $[0,2\pi )$.

Answer 1

Try to simplify the equation using identities of multiple angles and then use the formula of the general solution of $\cos x$.

Complete step by step solution:
Given equation is ${\sin ^2}x - {\cos ^2}x = 0$. Observe that the term on the left hand side is similar to a form of $\cos 2x$.
It is known that:
$\cos 2x$ $=$ ${\cos ^2}x - {\sin ^2}x$
$\Rightarrow {\sin ^2}x - {\cos ^2}x = - \cos 2x$
$\therefore {\sin ^2}x - {\cos ^2}x = 0$
$\Rightarrow - \cos 2x = 0$
$\Rightarrow \cos 2x = 0$.
Now, use the formula of general solution:
If $\cos \theta = 0$,
$\theta = \dfrac{{\left( {2n + 1} \right)\pi }}{2},n \in Z$
$\therefore \cos 2x = 0$
$\Rightarrow 2x = \dfrac{{\left( {2n + 1} \right)\pi }}{2}$
Divide both sides of the equation by $2$:
$\Rightarrow x = \dfrac{{\left( {2n + 1} \right)\pi }}{4}$
Putting $n = 0,1,2,3$; $x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}$
These are the solutions of $x$ in the interval $[0,2\pi )$.
Hence, x \in \left\\{ {\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}} \right\\}.

Note: Students must be thorough with all the trigonometric formulas and identities and must be vigilant about how they can be applied to simplify the question. The difference between the principal solution and a general solution must also be known. Principal solution is the solution that lies within the interval $\left[ {0,2\pi } \right]$. General solutions include all possible solutions of an equation. Also, note that the formula of a general solution can be used to determine the principal solutions.