Solving Trigonometric Equations, 3

Category: Trigonometry

"Published in Newark, California, USA"

Solve for the unknown angle for

Solution:

Consider the given equation

As you notice that the coefficients of Sin 2θ and Cos 2θ are not equal. If you will expand the above equation in order to convert the double angles into single angles, the above equation will be more complicated and it will be hard to simplify and solve the equation. Don't worry, we have a solution or technique to solve this kind of trigonometric equation. 

Draw a right triangle to represent the coefficients of Sin 2θ and Cos 2θ. The coefficient of Sin 2θ will be the adjacent side of a right triangle. The coefficient of Cos 2θ will be the opposite side of a right triangle. Solve for the hypotenuse and get the trigonometric functions as follows

Photo by Math Principles in Everyday Life

From the given equation

Divide both sides of the equation by the hypotenuse which is 2 as follows

Substitute the coefficients of Sin 2θ and Cos 2θ by their equivalent trigonometric functions as follows

 

but 

The above equation becomes

where n = number of revolutions.

Solving Trigonometric Equations, 2

Category: Trigonometry

"Published in Newark, California, USA"

Solve for the unknown angle for

Solution:

The first that we have to do is to reduce the higher angles in order to simplify the given equation. Consider the given equation above

Apply the Sum and Product of Two Angles Formula in order to reduce the higher angles as follows

Equate each factor to zero, we have

for

and

where n = number of revolutions.

for

and

and

where n = number of revolutions.

Word Problem - Average Problem

Category: Algebra

"Published in Newark, California, USA"

Lilian's average in 3 quizzes is 80%. If she made 5% more in the first than in the second and 8% less in the third, what grades did she get in the three quizzes?

Solution:

The given item in a word problem is the average of the three quizzes of Lilian with conditions of her quizzes and we have to find the grades of her quizzes as follows

Let    x = the score of her 2nd quiz
        x + 5% = the score of her 1st quiz
        x - 8% = the score of her 3rd quiz

            Average of three quizzes = ⅓(1st Quiz + 2nd Quiz + 3rd Quiz)

                          80% = ⅓ (x + 5% + x + x - 8%)

                                   3x - 3% = 240%

                                           3x = 240% + 3%

                                           3x = 243%

                                             x = 81%

Therefore

Lilian's grade in 1st Quiz = x + 5% = 81% + 5% = 86%

Lilian's grade in 2nd Quiz = x = 81%

Lilian's grade in 3rd Quiz = x - 8% = 81% - 8% = 73%