Question

Solve the expression and find the value of x.
$\sin x+\cos =1+\sin x\cos x$.

Answer 1

Hint: Square the given expression on both sides. Apply basic trigonometric identities and solve to get the value of x.

Complete step-by-step answer:

Given to us is the expression,
$\sin x+\cos =1+\sin x\cos x$
Let us square both LHS and RHS.
${{\left( \sin x+\cos \right)}^{2}}={{\left( 1+\sin x\cos x \right)}^{2}}$
We know ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
${{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=1+{{\sin }^{2}}x{{\cos }^{2}}x+2\sin x\cos x$
Let us cancel $2\sin x\cos x$ from both the sides.
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.

& \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1+{{\sin }^{2}}x{{\cos }^{2}}x \\\ & 1=1+{{\sin }^{2}}x{{\cos }^{2}}x \\\ \end{aligned}$$ Cancel out 1 from LHS and RHS. Hence we get, $${{\sin }^{2}}x{{\cos }^{2}}x=0$$ So we can write it as, $$\operatorname{sinx}=0$$ or $$\cos x=0$$. $$x={{\sin }^{-1}}0$$ and $$x={{\cos }^{-1}}0$$ From the trigonometric table we know $$\sin 0=0$$ or $$\cos 90=0$$. Thus for $$\sin x=0,x=n\pi $$ and for $$\cos x=0,x=\dfrac{\left( 2n+1 \right)\pi }{2},n\in N$$ Thus we got the value of x as $$n\pi $$ or $$\dfrac{\left( 2n+1 \right)\pi }{2},n\in N$$. Note: In the last step of the solution we must consider the condition that both the sin$x$ and cos$x$ can be zero, thereby we have to consider the angles of both because single alone happening zero will also make the complete multiplication zero.