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Question: How do you solve \(\sin x+\cos x=1\)?...

How do you solve sin⁡x+cos⁡x=1\sin x+\cos x=1?

Explanation

Solution

From the given question we have to solve the sin⁡x+cos⁡x=1\sin x+\cos x=1. To solve the above equation first we have to square the equation on both sides and then we have to proceed further. By simplifying the equation further we will get the values of the x. By this we can solve the above equation.

Complete step by step solution:
From the question given we have to solve the
⇒sin⁡x+cos⁡x=1\Rightarrow \sin x+\cos x=1
To solve the above question. First, we have to square the equation on both sides,
By squaring the equations on both sides, we will get,
⇒(sin⁡x+cos⁡x)2=(1)2\Rightarrow {{\left( \sin x+\cos x \right)}^{2}}={{\left( 1 \right)}^{2}}
By expanding the equation, we will get,
⇒sin⁡x2+cos⁡x2+2×sin⁡x×cos⁡x=1\Rightarrow \sin {{x}^{2}}+\cos {{x}^{2}}+2\times \sin x\times \cos x=1
As we know the trigonometric identities, the very familiar and well-known trigonometric identity is
⇒sin⁡x2+cos⁡x2=1\Rightarrow \sin {{x}^{2}}+\cos {{x}^{2}}=1
We have to use this identity in the above expression,
By using the above identity in the above expression, we will get,
⇒1+2×sin⁡x×cos⁡x=1\Rightarrow 1+2\times \sin x\times \cos x=1
Now we have to shift the one from left hand side to the right-hand side of the equation,
By shifting the one from left hand side of the equation to right hand side then we will get,
⇒2×sin⁡x×cos⁡x=1−1\Rightarrow 2\times \sin x\times \cos x=1-1
⇒2×sin⁡x×cos⁡x=0\Rightarrow 2\times \sin x\times \cos x=0
As we know that
⇒sin⁡2x=2×sin⁡x×cos⁡x\Rightarrow \sin 2x=2\times \sin x\times \cos x
By this we will get the above equation as
⇒sin⁡2x=0\Rightarrow \sin 2x=0
as we know that the general solutions of trigonometric equation,
⇒sin⁡2x=sin⁡2α\Rightarrow \sin 2x=\sin 2\alpha
⇒x=nπ±α\Rightarrow x=n\pi \pm \alpha
By this we get the solution of the given equation,
⇒2x=0,π\Rightarrow 2x=0,\pi
⇒x=2πn,π2+2πn\Rightarrow x=2\pi n,\dfrac{\pi }{2}+2\pi n
Where n is an integer.

Note: Student should know the basic formulas of trigonometric equations like,
⇒sin⁡2x+cos⁡2x=1 ⇒sec⁡2x−tan⁡2x=1 ⇒cos⁡ec2x−cot⁡2x=1 \begin{aligned} & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\\ & \Rightarrow {{\sec }^{2}}x-{{\tan }^{2}}x=1 \\\ & \Rightarrow \cos e{{c}^{2}}x-{{\cot }^{2}}x=1 \\\ \end{aligned}
Students should also know the general solutions of the trigonometric equations like,
⇒sin⁡θ=sin⁡α,θ=nπ+(−1)nα,where α∈[−π2,π2] ⇒cos⁡θ=cos⁡α,θ=2nπ±α,where α∈(0,π] ⇒tan⁡θ=tan⁡α,θ=nπ+α,where α∈(−π2,π2] \begin{aligned} & \Rightarrow \sin \theta =\sin \alpha ,\theta =n\pi +\left( -1 \right)n\alpha ,where\,\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] \\\ & \Rightarrow \cos \theta =\cos \alpha ,\theta =2n\pi \pm \alpha ,where\,\alpha \in \left( 0,\pi \right] \\\ & \Rightarrow \tan \theta =\tan \alpha ,\theta =n\pi +\alpha ,where\,\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] \\\ \end{aligned}