Question

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

Answer 1

Hint: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 $
Now divide each term of the equation by $ \cos x $
 $ \Rightarrow \dfrac{{\cos x + \sin x}}{{\cos x}} = \dfrac{0}{{\cos x}} $
By the trigonometry ratios $ \dfrac{1}{{\cos x}} $ can be written as $ \sec x $ . Using this trigonometry
ratio in the above equation we have
 $ \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
 $ \Rightarrow \left( {\cos x + \sin x} \right)\sec x = 0 $
Multiply $ \sec x $ to each term we have
 $ \Rightarrow \cos x \cdot \sec x + \sin x \cdot \sec x = 0 $
By the trigonometry ratios $ \sec x $ can be written as $ \dfrac{1}{{\cos x}} $ . Using this trigonometry
ratio in the above equation we have
 $ \Rightarrow \cos x \cdot \dfrac{1}{{\cos x}} + \sin x \cdot \dfrac{1}{{\cos x}} = 0 $
We can cancel $ \cos x $ in the first term of the LHS then we have
 $ \Rightarrow 1 + \dfrac{{\sin x}}{{\cos x}} = 0 $
By the trigonometry ratios we have $ \dfrac{{\sin x}}{{\cos x}} = \tan x $ , by using this we have
 $ \Rightarrow 1 + \tan x = 0 $
Add -1 to the both sides of the equation
 $ \Rightarrow 1 + \tan x - 1 = - 1 $
The +1 and -1 will cancels and we get
 $ \Rightarrow \tan x = - 1 $
To find the value of x we use inverse trigonometry to the above equation
Therefore, we have
 $ \Rightarrow x = {\tan ^{ - 1}}( - 1) $
The exact value of $ {\tan ^{ - 1}}( - 1) $ is $ - \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 $ \pi $
 $
\Rightarrow x = - \dfrac{\pi }{4} + \pi \\
\Rightarrow x = \dfrac{{3\pi }}{4} \\
 $
The period of $ \tan x $ functions is $ \pi $ so values will repeat every $ \pi $ radians in both directions.
 $ x = \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.