Question

Solve the following equations.
$si{n}^{3}x-co{s}^{3}x=1+sinxcosx.$

Answer 1

$si{n}^{3}x-co{s}^{3}x=1+sinxcosx.$

Solution:- We are given that,

$si{n}^{3}x-co{s}^{3}x-1+sinxcosx$

$\to$ now, we know that $a^3-b^3=(a-b)(a^2+ab+b^2)$

So,

$(sinx-cosx)(si{n}^{2}x+sinxcosx+co{s}^{2}x)=1+sinxcosx$

$(sinx-cosx)(1+sinxcosx)-(1+sinxcosx)=0$

$(si{n}^{2}x+co{s}^{2}x=1)$

$(1+sinxcosx)\left[\right(sinx-cosx)-1]=0$

So,

$1+sinxcosx=0$ or $sinx-cosx=1$

$2+2sinxcosx=0$ or $1/\sqrt{2}(sinx-cosx)=1/\sqrt{2}$

$(\because$ multiply with 2) or $(\because$ multiply by $1/\sqrt{2})$

$2+2\sin 2x=0$ or $1\sqrt{2}sinx-1/2cosx=1/\sqrt{2}$

$sin2x=-2$ or $sinxcos\frac{\pi }{4}-cosxsin\frac{\pi }{4}=1/\sqrt{2}$

which is not possible

$(\because -1\le sin\theta \le 1)$ or $sin(x-\frac{pi}{4})\frac{1}{\sqrt{2}}$

$\to$ now, the general solution of

$sin\theta =asin\alpha ,aϵ[-1,1]$

$\alpha ϵ[-\pi /2,\pi /2]$

$sin\alpha =a=\frac{1}{\sqrt{2}}$

$\alpha =si{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}.$

is, $\theta =k\pi +(-1{)}^k\alpha ,kϵz$

$\theta =k\pi +(-1{)}^k\frac{\pi }{4},kϵz$

now here, $sin(x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}$

So, $x-\frac{\pi }{4}=k\pi +(-1{)}^k\frac{\pi }{4}$

$x=k\pi +(-1{)}^k\frac{\pi }{4}+\frac{\pi }{4},kϵz$

The required set of solution of x

$\{k\pi +(-1{)}^k\frac{\pi }{4}+\frac{\pi }{4}/kϵz\}$