Solve. Find all solutions in [0,2).5 secx cotx+5 secx+ cotx+1=0

1 answer:

5 0

$\begin{gathered} 5sec(x)cot(x)+5sec(x)+cot(x)+1=0 \\ Factoring \\ (cot(x)+1)(5sec(x)+1)=0 \\ cot(x)+1=0 \\ cot(x)=-1 \\ x=cot^{-1}(1) \\ x=\frac{3\pi}{4} \\ \\ 5sec(x)+1=0 \\ 5sec(x)=-1 \\ sec(x)=\frac{-1}{5},\text{ there is no solution} \\ Hence \\ All\text{ solution are }\frac{3\pi}{4}+\pi n,\text{ where n=1,2,3...} \\ \\ \end{gathered}$

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What binomial do you have to add to the polynomial 2x^2+3y^2-5xy+1 to get the following?

Contact [7]

Binomial $-3y^{2}+5xy$ when added to polynomial $2x^{2} +3y^{2} -5xy+1$ gives polynomial that does not contain the variable y .

<h3>What is binomial?</h3>

A mathematical expression consisting of two terms connected by a plus sign or minus sign .

Example: x + 2 is a binomial, where x and 2 are two separate terms.

<h3>What is polynomial?</h3>

A polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer

According to the question

polynomial $2x^{2} +3y^{2} -5xy+1$

when added with $-3y^{2}+5xy$

= $2x^{2}+1$ (as $3y^{2} -5xy$ got cancelled )