SOLVING TRIGONOMETRIC EQUATIONS
About "Solving Trigonometric Equations"
Solving Trigonometric Equations :
The equations containing trigonometric functions of unknown angles are known as trigonometric equations. A solution of trigonometric equation is the value of unknown angle that satisfies the equation.
General Solution :
The solution of a trigonometric equation giving all the admissible values obtained with the help of periodicity of a trigonometric function is called the general solution of the equation.
Principal Solution
The smallest numerical value of unknown angle satisfying the equation in the interval [0, 2π] (or) [−π, π] is called a principal solution.
Principal value of sine function lies in the interval
[−π/2, π/2]
and hence lies in I quadrant or IV quadrant.
Principal value of cosine function is in
[0, π]
and hence in I quadrant or II quadrant.
Principal value of tangent function is in
(-π/2, π/2)
and hence in I quadrant or IV quadrant.
How to Solve Trigonometric Equations step by step
Question 1 :
Solve the following equations for which solutions lies in the interval 0° ≤ θ < 360°
(i) sin4 x = sin2 x
Solution :
sin4 x = sin2 x
sin4 x - sin2 x = 0
sin2 x (sin2 x - 1) = 0
|
sin2 x = 0 x = sin-1(0) x = 0, π, 2π, ...... |
sin2 x - 1 = 0 sin2 x = 1 sin x = √1 x = sin-1(1) (or) x = sin-1 (-1) x = π/2 and x = 3π/2 |
Since we choose the values between 0 to 360, the solution will be {0, π/2, π, 3π/2}.
(ii) 2 cos2 x + 1 = −3 cos x
Solution :
2 cos2 x + 1 = −3 cos x
2 cos2 x + 1 + 3 cos x = 0
Let t = cos x
2t2 + 3t + 1 = 0
By factoring the quadratic equation, we get
(2t + 1)(t + 1) = 0
|
2t + 1 = 0 2t = -1 t = -1/2 cos x = -1/2 |
t + 1 = 0 t = -1 cos x = -1 |
For cos x = -1/2
For negative values of cos, we have to select the angle from 2nd and 3rd quadrants.
θ = π - a and π + a
x = π - (π/3) x = π + (π/3)
x = 2π/3 x = 4π/3
cos x = -1
x = π
Hence the solution is {π, 2π/3, 4π/3}.
(iii) 2 sin2 x+1 = 3sinx
Solution :
2 sin2 x+1 = 3sinx
Let t = sin x
2 sin2 x - 3sinx + 1 = 0
2t2 - 3t + 1 = 0
By factoring the quadratic equation, we get
(2t - 1)(t - 1) = 0
|
2t - 1 = 0 2t = 1 t = 1/2 sin x = -1/2 |
t - 1 = 0 t = 1 sin x = 1 |
For sin x = 1/2
x = π/6
For positive value of sin, we have to select angles from 2nd quadrant
x = π - a
x = π - (π/6)
x = 5π/6
For sin x = 1
x = π/2
Hence the required solution is { π/2, π/6, 5π/6}
(iv) cos 2x = 1 - 3 sin x
Solution :
cos 2x = 1 - 3 sin x
1 - 2 sin2x = 1 - 3 sin x
2sin2x - 3 sin x + 1 - 1 = 0
2sin2x - 3 sin x = 0
sin x (2 sin x - 3) = 0
sin x = 0 2 sin x - 3 = 0
x = sin-1(0) 2 sin x = 3
x = 0, π sin x = 3/2
Hence the solution is {0, π}.
After having gone through the stuff given above, we hope that the students would have understood, "Solving Trigonometric Equations"
Apart from the stuff given in "Solving Trigonometric Equations", if you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Markup and markdown word problems
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Word problems on sets and venn diagrams
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
