Give answers of angles in radians in terms of [tex]$\pi$[/tex].
QUESTION 1
Find the exact solution for [tex]$x$[/tex], if there is no solution write no solution:
1. [tex]$1.15 e^{3 x}-4=6$[/tex]
2. [tex]$1.2 e^{2 x}-e^x-110=0$[/tex]
3. [tex]$1.3 \ln \left(e^x\right)+e^{(\ln 5)}=25$[/tex]
4. [tex]$1.4 \log _8(7)+\log _8(-4 x)=\log _8(5)$[/tex]
5. [tex]$1.5 \sin ^2 x-\cos ^2 x-\sin x=0$[/tex] on [tex]$[0,2 \pi)$[/tex].
6. [tex]$x=\sin \left(70^{\circ}\right) \cos \left(25^{\circ}\right)-\cos \left(70^{\circ}\right) \sin \left(25^{\circ}\right)$[/tex]
QUESTION 2
Show that the expression [tex]$\frac{\left(1-\cos ^2 x\right)\left(1+\cos ^2 x\right)}{\cos ^2 x}$[/tex] is equivalent to [tex]$\sin ^2 x+\sec ^2$[/tex]
Answer (1)
Solve the exponential equation in 1.1 by isolating the exponential term and taking the natural logarithm: x = 3 l n 2 .
Solve the exponential equation in 1.2 by substituting u = e x and solving the resulting quadratic equation: x = ln 11 .
Solve the logarithmic equation in 1.3 using logarithm properties: x = 20 .
Solve the logarithmic equation in 1.4 using logarithm properties: x = − 28 5 .
Solve the trigonometric equation in 1.5 by using trigonometric identities and solving the resulting quadratic equation: x = 2 π , 6 7 π , 6 11 π .
Solve the trigonometric equation in 1.6 using the sine subtraction formula: x = 2 2 .
Prove the trigonometric identity in 2 by simplifying the left side and using trigonometric identities to show it is equivalent to the right side. x = 3 ln 2 , ln 11 , 20 , − 28 5 , 2 π , 6 7 π , 6 11 π , 2 2
Explanation
Problem Analysis We are tasked with solving several equations for x and proving a trigonometric identity. Let's tackle each part step by step, expressing angles in radians in terms of π where applicable.
Solving 1.1 1.1: Solve 5 e 3 x − 4 = 6 . First, isolate the exponential term: 5 e 3 x = 10 e 3 x = 2 Now, take the natural logarithm of both sides: 3 x = ln 2 x = 3 ln 2
Solving 1.2 1.2: Solve e 2 x − e x − 110 = 0 . Let u = e x . Then the equation becomes: u 2 − u − 110 = 0 Factoring the quadratic equation, we have: ( u − 11 ) ( u + 10 ) = 0 So, u = 11 or u = − 10 . Since u = e x , u must be positive, so u = 11 .
e x = 11 x = ln 11 ≈ 2.3979
Solving 1.3 1.3: Solve ln ( e x ) + e l n 5 = 25 . Using logarithm properties, we have: x + 5 = 25 x = 20
Solving 1.4 1.4: Solve lo g 8 ( 7 ) + lo g 8 ( − 4 x ) = lo g 8 ( 5 ) . Using logarithm properties, we have: lo g 8 ( 7 ( − 4 x )) = lo g 8 ( 5 ) 7 ( − 4 x ) = 5 − 28 x = 5 x = − 28 5
Solving 1.5 1.5: Solve sin 2 x − cos 2 x − sin x = 0 on [ 0 , 2 π ) . Using the identity cos 2 x = 1 − sin 2 x , we have: sin 2 x − ( 1 − sin 2 x ) − sin x = 0 2 sin 2 x − 1 − sin x = 0 2 sin 2 x − sin x − 1 = 0 Let u = sin x . Then the equation becomes: 2 u 2 − u − 1 = 0 ( 2 u + 1 ) ( u − 1 ) = 0 So, u = 1 or u = − 2 1 .
If sin x = 1 , then x = 2 π .
If sin x = − 2 1 , then x = 6 7 π or x = 6 11 π .
Therefore, the solutions are x = 2 π , 6 7 π , 6 11 π .
Solving 1.6 1.6: Solve x = sin ( 7 0 ∘ ) cos ( 2 5 ∘ ) − cos ( 7 0 ∘ ) sin ( 2 5 ∘ ) . Using the sine subtraction formula, we have: x = sin ( 7 0 ∘ − 2 5 ∘ ) = sin ( 4 5 ∘ ) = 2 2 Converting to radians, 4 5 ∘ = 4 π . Thus, x = sin ( 4 π ) = 2 2 .
Proving the Trigonometric Identity 2: Show that c o s 2 x ( 1 − c o s 2 x ) ( 1 + c o s 2 x ) is equivalent to sin 2 x + sec 2 x − 1 .
Starting with the left side: cos 2 x ( 1 − cos 2 x ) ( 1 + cos 2 x ) = cos 2 x sin 2 x ( 1 + cos 2 x ) = cos 2 x sin 2 x + cos 2 x sin 2 x cos 2 x = tan 2 x + sin 2 x Using the identity tan 2 x + 1 = sec 2 x , we have tan 2 x = sec 2 x − 1 . Substituting this into the expression: sec 2 x − 1 + sin 2 x = sin 2 x + sec 2 x − 1 Thus, the expression is equivalent to sin 2 x + sec 2 x − 1 .
Final Answer The solutions for the equations are: 1.1: x = 3 l n 2 1.2: x = ln 11 1.3: x = 20 1.4: x = − 28 5 1.5: x = 2 π , 6 7 π , 6 11 π 1.6: x = 2 2 The trigonometric identity is proven.
Examples
Understanding trigonometric identities and solving equations is crucial in fields like physics and engineering. For instance, when analyzing the motion of a pendulum, we use trigonometric functions to describe its position and velocity over time. Similarly, in electrical engineering, alternating current (AC) circuits are modeled using sinusoidal functions, and solving equations involving these functions helps engineers design and optimize circuits. These mathematical tools enable us to predict and control real-world phenomena.
