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Question: Solve \(\sin \left( x \right) + \cos \left( x \right) = 1\) ....

Solve sin(x)+cos(x)=1\sin \left( x \right) + \cos \left( x \right) = 1 .

Explanation

Solution

The given question involves solving a trigonometric equation and finding the value of angle x that satisfies the given equation. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities.

Complete step by step answer:
The given problem requires us to solve the trigonometric equation sin(x)+cos(x)=1\sin \left( x \right) + \cos \left( x \right) = 1 .
The given trigonometric equation can be solved by condensing both the terms into a single compound angle sine or cosine formula.
Dividing both sides of the equation by 12\dfrac{1}{{\sqrt 2 }} to form a condensed trigonometric formula, we get,
sin(x)2+cos(x)2=12\dfrac{{\sin \left( x \right)}}{{\sqrt 2 }} + \dfrac{{\cos \left( x \right)}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}
We know that sin(π4)=cos(π4)=12\sin \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}, we get,
=sin(x)cos(π4)+cos(x)sin(π4)=12= \sin \left( x \right)\cos \left( {\dfrac{\pi }{4}} \right) + \cos \left( x \right)\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}
Using the compound formulae of sine and cosine, sin(A+B)=sinAcosB+cosAsinB\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B and cos(A+B)=cosAcosBsinAsinB\cos (A + B) = \cos A\cos B - \sin A\sin B, we get,
=sin(x+π4)=sin(π4)= \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right)
Now, we know that general solution for the equation sin(θ)=sin(ϕ)\sin \left( \theta \right) = \sin \left( \phi \right) is θ=nπ+(1)nϕ\theta = n\pi + {\left( { - 1} \right)^n}\phi , where n is any integer.
=(x+π4)=nπ+(1)n(π4)= \left( {x + \dfrac{\pi }{4}} \right) = n\pi + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{4}} \right), where n is an integer.
Taking (π4)\left( {\dfrac{\pi }{4}} \right) to right side of the equation, we get,
=x=nπ+(1)n(π4)π4= x = n\pi + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{4}} \right) - \dfrac{\pi }{4} , where n is any integer.
So, solution of the given equation sin(x)+cos(x)=1\sin \left( x \right) + \cos \left( x \right) = 1 is: x=nπ+(1)n(π4)π4x = n\pi + {\left( { - 1} \right)^n}\left( {\dfrac{\pi }{4}} \right) - \dfrac{\pi }{4}, where n is any integer.

Note: Such trigonometric equations can be solved by various methods by applying suitable trigonometric identities and formulae. The general solution of a given trigonometric solution may differ in form, but actually represents the correct solutions. The different forms of general equations are interconvertible into each other. For solving such types of questions where we have to solve trigonometric equations, we need to have basic knowledge of algebraic rules and identities as well as a strong grip on trigonometric formulae and identities.