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Question: How do you solve \( \cos x + \sin x = 0 \) ?...

How do you solve cos⁡x+sin⁡x=0\cos x + \sin x = 0 ?

Explanation

Solution

Here the question is related to the trigonometry, we use the trigonometry ratios and we are to solve this question. In this question we can find the value of x such that it satisfies the given question. The value of x is determined with the help of angle of trigonometry ratios table.

Complete step by step explanation:
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine and tan.
Now consider the given question
cos⁡x+sin⁡x=0\cos x + \sin x = 0
Now divide each term of the equation by cos⁡x\cos x
⇒cos⁡x+sin⁡xcos⁡x=0cos⁡x\Rightarrow \dfrac{{\cos x + \sin x}}{{\cos x}} = \dfrac{0}{{\cos x}}
By the trigonometry ratios 1cos⁡x\dfrac{1}{{\cos x}} can be written as sec⁡x\sec x . Using this trigonometry
ratio in the above equation we have
⇒(cos⁡x+sin⁡x)sec⁡x=0⋅sec⁡x\Rightarrow \left( {\cos x + \sin x} \right)\sec x = 0 \cdot \sec x
Any number multiplied by zero then the solution is also zero so we have
⇒(cos⁡x+sin⁡x)sec⁡x=0\Rightarrow \left( {\cos x + \sin x} \right)\sec x = 0
Multiply sec⁡x\sec x to each term we have
⇒cos⁡x⋅sec⁡x+sin⁡x⋅sec⁡x=0\Rightarrow \cos x \cdot \sec x + \sin x \cdot \sec x = 0
By the trigonometry ratios sec⁡x\sec x can be written as 1cos⁡x\dfrac{1}{{\cos x}} . Using this trigonometry
ratio in the above equation we have
⇒cos⁡x⋅1cos⁡x+sin⁡x⋅1cos⁡x=0\Rightarrow \cos x \cdot \dfrac{1}{{\cos x}} + \sin x \cdot \dfrac{1}{{\cos x}} = 0
We can cancel cos⁡x\cos x in the first term of the LHS then we have
⇒1+sin⁡xcos⁡x=0\Rightarrow 1 + \dfrac{{\sin x}}{{\cos x}} = 0
By the trigonometry ratios we have sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x , by using this we have
⇒1+tan⁡x=0\Rightarrow 1 + \tan x = 0
Add -1 to the both sides of the equation
⇒1+tan⁡x−1=−1\Rightarrow 1 + \tan x - 1 = - 1
The +1 and -1 will cancels and we get
⇒tan⁡x=−1\Rightarrow \tan x = - 1
To find the value of x we use inverse trigonometry to the above equation
Therefore, we have
⇒x=tan⁡−1(−1)\Rightarrow x = {\tan ^{ - 1}}( - 1)
The exact value of tan⁡−1(−1){\tan ^{ - 1}}( - 1) is −π4- \dfrac{\pi }{4}

As we know about the ASTC rule the tan is negative in the second quadrant and the fourth quadrant.
To find the solution add the reference angle and π\pi
⇒x=−π4+π ⇒x=3π4  \Rightarrow x = - \dfrac{\pi }{4} + \pi \\\ \Rightarrow x = \dfrac{{3\pi }}{4} \\\
The period of tan⁡x\tan x functions is π\pi so values will repeat every π\pi radians in both directions.
x=3π4+πnx = \dfrac{{3\pi }}{4} + \pi n , for any integer n.

Note: Since we can say at which value of x the addition of cosine and sine will be zero. The ASTC rule defined as all sine tan cosine this explains all trigonometry ratios are positive in the first quadrant. Sine trigonometry ratio is positive in the second quadrant. tan trigonometry ratio is positive in the third quadrant and the cosine trigonometry ratio is positive in the fourth quadrant.