Question

Solve $\sin x+\cos x=1+\sin x\cos x$ , then the solution of x is $2n\pi \pm \dfrac{\pi }{4}$ . If true then enter 1 and if false then enter 0.

Answer 1

Hint: First of all, square both sides of the equation $\sin x+\cos x=1+\sin x\cos x$ . Now, use the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and simplify the equation $({{\sin }^{2}}x+{{\cos }^{2}}x=1+{{\sin }^{2}}x{{\cos }^{2}}x)$ . Now, we have to get the solution of $\sin x=0$ and $\cos x=0$ . We know that $\sin 0=0$ , $\sin \pi =0$ , and $\sin 2\pi =0$ . So, $\sin x=\sin n\pi$ . We know that $\cos \dfrac{\pi }{2}=0$ , $\cos \left( -\dfrac{\pi }{2} \right)=0$ , and $\cos \left( \dfrac{3\pi }{2} \right)=0$ . So, $\cos x=\cos \left( \dfrac{3\pi }{2} \right)$ . Now, solve it further and get the value of x.

Complete step-by-step answer:
According to the question, our given expression is $\sin x+\cos x=1+\sin x\cos x$ ……………(1)
We have to get the solutions of x.
Now, squaring on both LHS and RHS of the equation (1), we get
$\sin x+\cos x=1+\sin x\cos x$
$\Rightarrow {{(\sin x+\cos x)}^{2}}={{(1+\sin x\cos x)}^{2}}$ ……………………………(2)
We know the formula, ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ . We can expand equation (2) with the help of the formula.
Using the formula and expanding equation (2), we get
$\Rightarrow {{(\sin x+\cos x)}^{2}}={{(1+\sin x\cos x)}^{2}}$

& \Rightarrow {{(\sin x+\cos x)}^{2}}={{(1+\sin x\cos x)}^{2}} \\\ & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=1+{{\sin }^{2}}x{{\cos }^{2}}x+2\sin x\cos x \\\ \end{aligned}$$ $$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1+{{\sin }^{2}}x{{\cos }^{2}}x$$ ………………………(3) We know the identity, $${{\sin }^{2}}x+{{\cos }^{2}}x=1$$ ……………(4) Using equation (4) and replacing $$({{\sin }^{2}}x+{{\cos }^{2}}x)$$ by 1 in equation (3), we get $$\begin{aligned} & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1+{{\sin }^{2}}x{{\cos }^{2}}x \\\ & \Rightarrow 1=1+{{\sin }^{2}}x{{\cos }^{2}}x \\\ \end{aligned}$$ $$\Rightarrow 0={{\sin }^{2}}x{{\cos }^{2}}x$$ $$\Rightarrow 0={{\sin }^{2}}x{{\cos }^{2}}x$$ $$\Rightarrow 0=\sin x\cos x$$ So, either $$\sin x=0$$ or $$\cos x=0$$ ………………..(5) Let us solve $$\sin x=0$$ . We know that, $$\sin 0=0$$ …………….(6) From equation (5) and equation (6), we get $$\begin{aligned} & \Rightarrow \sin x=\sin 0 \\\ & \Rightarrow x=0 \\\ \end{aligned}$$ We know that, $$\sin \pi =0$$ ……………...(7) From equation (5) and equation (7), we get $$\begin{aligned} & \Rightarrow \sin x=\sin \pi \\\ & \Rightarrow x=\pi \\\ \end{aligned}$$ We know that, $$\sin 2\pi =0$$ …………...(8) From equation (5) and equation (8), we get $$\begin{aligned} & \Rightarrow \sin x=\sin 2\pi \\\ & \Rightarrow x=2\pi \\\ \end{aligned}$$ We can see that x is a multiple of $$\pi $$ . So, $$x=n\pi $$ ……………………….(9) Let us solve $$\cos x=0$$ . We know that, $$\cos \dfrac{\pi }{2}=0$$ …………….(10) From equation (5) and equation (10), we get $$\begin{aligned} & \Rightarrow \cos x=\cos \dfrac{\pi }{2} \\\ & \Rightarrow x=\dfrac{\pi }{2} \\\ \end{aligned}$$ We know that, $$\cos \dfrac{3\pi }{2}=0$$ ……………...(11) From equation (5) and equation (11), we get $$\begin{aligned} & \Rightarrow \cos x=\cos \dfrac{3\pi }{2} \\\ & \Rightarrow x=\dfrac{3\pi }{2} \\\ \end{aligned}$$ We know that, $$\cos \left( -\dfrac{\pi }{2} \right)=0$$ …………...(12) From equation (5) and equation (12), we get $$\begin{aligned} & \Rightarrow \cos x=\cos \left( -\dfrac{\pi }{2} \right) \\\ & \Rightarrow x=\dfrac{-\pi }{2} \\\ \end{aligned}$$ Now, we have values of x which are $$-\dfrac{\pi }{2},\dfrac{\pi }{2},and\dfrac{3\pi }{2}$$ . So, $$x=n\pi \pm \dfrac{\pi }{2}$$ ……………………(13) From equation (9) and equation (13), we have the solutions for x which are $$x=n\pi $$ and $$x=n\pi \pm \dfrac{\pi }{2}$$ . So, $$x=2n\pi \pm \dfrac{\pi }{4}$$ is not the solution of x. It is given that we have to enter 1 if true and if false then we have to enter 0. Thus, we have to enter 0. Note: In this question, one may think to use the formula $$\sin 2x=2\sin x\cos x$$ to solve the equation $$({{\sin }^{2}}x{{\cos }^{2}}x=0)$$ . Using the formula, $$\sin 2x=2\sin x\cos x$$ we can replace $${{\sin }^{2}}x{{\cos }^{2}}x$$ by $$\dfrac{{{\sin }^{2}}2x}{4}$$ . $$\begin{aligned} & \dfrac{{{\sin }^{2}}2x}{4}=0 \\\ & \Rightarrow \sin 2x=0 \\\ \end{aligned}$$ Now, solving $$\sin 2x=0$$ , $$\begin{aligned} & \sin 2x=\sin n\pi \\\ & \Rightarrow 2x=n\pi \\\ & \Rightarrow x=\dfrac{n\pi }{2} \\\ \end{aligned}$$ Here, we have got only $$x=\dfrac{n\pi }{2}$$ . Using this formula makes us lose the solution of $$\cos x=0$$ . So, we don’t have to use this formula here.