TRIGONOMETRIC EQUATIONS AND SOLVING METHODS

Аннотация

This in the article trigonometric equations solution non-standard methods about information given . Including both sides of the equation side one kind trigonometric to the function multiplication , both​ towards one odd number or one kind trigonometric function add and subtraction , proportion , mathematics analysis of elements , vectors scalar from the product useabout information given andvisols undress shown. Key words :Function determination domain of the function limitation property , proportion , vector,scalar multiplication,number inequality ,equations system ,equations union . Trigonometric equations to the appearance looking at solution one how much methods available . These instead of to put , to rationalize instead of substitutions , trigonometric equations different ways of solving private cases , artificial form from substitutions using trigonometric equations solution and etc. Some​ in cases given we know the equations methods with solution much complicated It will be . equations solution non-standard to the methods stop Let's go . Artificial form requiring replacement​ trigonometric equations in solutionfollowing from methods is used . Trigonometry algebra and geometry in sciences important importance has is , is integrative to the feature have ​ In this sense trigonometric equations solutions collection generalization and them choice in students difficulty gives birth to . Because students two variable linear equations solution methods and their application opportunities about enough for information has not to be possible . Two variable linear equations solution methods school mathematics in science deep It is also known that it cannot be studied . But specialized schools and high schools mathematics study in programs two variable linear equations occurs and they Diophantus equations (undefined equations ) topicsin thesection isstudied . Two variable linear Diophantus equations solution technique trigonometric equation roots collection one how many when general roots the set separate in receiving application possible.Because selectively taken general roots set usually given equation solution as a result found all roots total satisfied for they two variable linear Diophantus equations roots set that obvious will be . Especially trigonometric equations given in the field when given roots collection separate in receiving this obvious It seems like this . in cases trigonometric equation roots only given to the sectors relevant those who were separate required to obtain and in this many in cases choice through results is taken . But choice and intuitive considerations always reliable​ expected the results It doesn't give . That's why for Diophantus equations and trigonometric of equations mutual relatedness study mathematics internal integration to strengthen service does .Mutually relevance trigonometric of equations solutions set first ofall arithmeticprogression organization tobe ablewith explained . page1770 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue05,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai Trigonometric issues in solving , theirgeneral solutions find for arithmetic progressions in the form of written roots set compare need will be , because they either of the matter solution,oronthecontrary tobe possible. Let's assume first x =+ 15kprogression k- hadi second x = +3 7mprogression m - k m to the extent equal let it be . k and m parameter how in values 1 5 + = + k 3 7 m equality appropriate that we determine . k and m parameter so value find must arithmetic progressions found numbered values equal Let . As a result following whole with coefficient two variable linear equation harvest becomes : 5 k − 7 m = 2. The equation whole numbers in the collection We solve the equation . variable in front of two from the coefficient we get the most small to value has that happened we choose ( this is 5) and on the left second the limit is 7 m = 5 m + 2 m such as we write and 5 k − − 5 m 2 m − = + 2 5 k 5 m 2 2 m . The left side of the equation is 5 submissiveness for right side is also 5 division necessary : 2 m + = 2 5 s , sZ . Thus , 2 m − =−5s 2from theequation m ands unknown wholenumbers as abovewe willfind . Last of the equation variables in front of two from the coefficient we get the most small to value has what happened we choose ( this is 2) and on the left second term 2m − − =− 4σs 2 2 m − = − 4 ss in 2 forms We write . The left side of the equation is divided by 2 . multiple , therefore for s − 2 is also 2 multiple to be we need : s − = 2 2 p , pZ .So as s = + 2 p 2from the equation unknown s and p whole numbers We will find . From the unknowns one of coefficient 1​ equal was equation to be taken with solution process finally So ,s =+ 2p 2​ equation pof optionalwhole invalues will beaninteger . So as first progression k = + 7 p6 in number present x = + = + 1 5 k 1 5(7p + = 6) 35 p