13. [tex]\sqrt{3} \cos x + \cot^2 x = \frac{\sin^3 x + 1}{\sin^2 x}[/tex]
14. [tex]4 \sin^2 x = \sin 2x + 2 \sin x + 5 \cos x + 5[/tex]
Answer (1)
To solve the given equations, let's break down each one step by step:
Equation 13: 3 cos x + cot 2 x = sin 2 x sin 3 x + 1
First, we need to simplify the expression on the right-hand side. By dividing each term in the numerator by sin 2 x , the equation becomes:
sin 2 x sin 3 x + sin 2 x 1 = sin x + csc 2 x
Substitute cot 2 x = csc 2 x − 1 into the left-hand side:
3 cos x + csc 2 x − 1 = sin x + csc 2 x
Now, cancel csc 2 x from both sides:
3 cos x − 1 = sin x
This is the simplified form of the first equation.
Equation 14: 4 sin 2 x = sin 2 x + 2 sin x + 5 cos x + 5
We know that sin 2 x = 2 sin x cos x . Substitute this into the equation:
4 sin 2 x = 2 sin x cos x + 2 sin x + 5 cos x + 5
To simplify, group all terms containing sin x together:
4 sin 2 x − 2 sin x cos x − 2 sin x = 5 cos x + 5
This equation can be simplified by further manipulation or solved for specific values of x where both equations hold true. However, further solving these trigonometric equations generally involves either analytical solutions or computational methods like graphing or using specific trigonometric values.
These equations involve trigonometric identities and relationships, making them a typical exercise in high school trigonometry, particularly in solving trigonometric equations. Understanding how to manipulate these equations using identities like sin 2 x + cos 2 x = 1 , csc x = s i n x 1 , and cot x = s i n x c o s x is crucial.
