1st Year Math Notes Chapter 14
Important Complete 1st Year Math Notes Chapter 14 written by Professor Asad Khalid Suib. These notes are very helpful in the preparation of Solutions of Trigonometric Equations for the students of Mathematics of the Intermediate and these are according to the paper patterns of all Punjab boards.
Summary and Contents:
Topics which are discussed in the notes are given below:
- Our comprehensive 1st Year Math Notes Chapter 14 will ensure you're fully prepared for your exams.
- Solution of the chapter number 14 in the 1st year of the Punjab Textbook Boards.
- Find the quadrant value of trigonometric functions.
- sin 2x is +ve in I and II Quadrants with the reference angle
- The Equations, containing at least one trigonometric function, are called Trigonometric Equations, e.g., each of the following is a trigonometric equation: sin , Sec tan and sin sec 1 Trigonometric equations have an ininite number of solutions due to the periodicity of the trigonometric functions. For examplemIf sin then 0, , 2 ,... qq q = =±± which can be written as q =∈n nZ , where . In solving trigonometric equations, irst ind the solution over the interval whose length is equal to its period and then ind the general solution as explained in the following examples:
- Solution of General Trigonometric Equations: When a trigonometric equation contains more than one trigonometric functions, trigonometric identities and algebraic formulae are used to transform such trigonometric equation to an equivalent equation that contains only one trigonometric function. The method is illustrated in the following solved examples:
- Note: In solving the equations of the form sin kx = c, we irst ind the solution pf sin u = c (where kx = w) and then required solution is obtained by dividing each term of this solution set by k.
- Sometimes it is necessary to square both sides of a trigonometric equation. In such a case, extaneous roots can occur which are to be discarded. So each value of x must be checked by substituting it in the given equation. For example, x = 2 is an equation having a root 2. On squaring we get x 2 - 4 which gives two roots 2 and -2. But the root -2 does not satisfy the equation x = 2. Therefore, -2 is an extaneous root.
- Since cos x is -v e in II and III Quadrants with the reference angle x = pi /3.
- Now both csc x and cot x are not defined for x = 0 and x = 2.
- Solve the equation: csc3x cot x .
- Since cos x is -v e in II and III Quadrants with the reference angle x = pi / 3.
- Now x = 4pi / 3 does not satisfy the given equation (i).
- x = 4pi / 3 is not admissible and so x = 2pi / 3 is the only solution.
- Since 2p is the period of cos x.
- Now both csc x and cot x are not defined for x = 0 and x = 2 ∴ x = 0 and x = 2 are not admissible.
- Find the solutions of the following equations which lie in [0, 2pi].
- Solve the following trigonometric equations: (i) tan2θ = 1/3.
- Find the values of θ satisfying the following equations: tan2θ - secθ -1 = 0.
- Find the solution sets of the following equations: 8. 4 sin2θ - 8cosθ + 1 = 0
- Find the solution sets of the following equations:
- 8. 4 sin2q - 8cosq + 1 = 0
- 9. 3 tan sec 1 0 x x - -=
- 10. cos 2x = sin 3x [Hint: sin3x = 3sinx - 4sin3x
- 11. sec 3q= secq
- 12. tan 2q + cotq = 0
- 13. sin 2x + sinx = 0
- 14. sin 4x - sin 2x = cos 3x
- 15. sin x + cos 3x = cos 5x
- 16. sin 3 x + sin 2x + sin x = 0
- 17. sin 7x - sin x = sin 3 x