Answer :
Answer:
[tex]\textsf{1)}\quad \theta=\dfrac{\pi}{3}+2k\pi,\;\;\theta=\dfrac{5\pi}{3}+2k\pi[/tex]
[tex]\textsf{2)}\quad \theta=0.34+2k\pi,\;\;\theta= 2.80+2k\pi[/tex]
[tex]\textsf{3)}\quad \theta=\dfrac{3\pi}{2}+2k\pi[/tex]
[tex]\textsf{4)}\quad \theta=\dfrac{\pi}{2}+2k\pi,\;\;\theta=\dfrac{3\pi}{2}+2k\pi[/tex]
Step-by-step explanation:
Question 1
Given trigonometric equation:
[tex]4 \cos^2\theta - 4 \cos \theta +1=0[/tex]
Rewrite the left side of the equation as:
[tex](2 \cos \theta)^2 - 2 \cdot 2\cos \theta +1^2=0[/tex]
Apply the Perfect Square Formula (a - b)² = a² - 2ab + b², where a = 2 cos θ and b = 1:
[tex](2\cos \theta -1)^2=0[/tex]
Now, square root both sides:
[tex]2\cos\theta-1=0[/tex]
Solve for cos θ:
[tex]2\cos\theta-1+1=0+1\\\\\\2\cos\theta=1\\\\\\\cos \theta=\dfrac{1}{2}[/tex]
According to the unit circle, cos θ equals 1/2 when θ = π/3 and θ = 5π/3. Therefore:
[tex]\theta=\dfrac{\pi}{3}+2\pi n,\;\;\theta=\dfrac{5\pi}{3}+2\pi n[/tex]
As the cosine function repeats itself every 2π radians, this means that if we add or subtract 2π to any solution, we get another valid solution. So, after finding the initial solutions θ = π/3 and θ = 5π/3, we add 2πn to account for all possible repetitions, where n is any integer. This ensures that all possible angles that satisfy the equation are covered.
In this case, you are using the variable 'k' instead of 'n' for any integer, and so are adding 2kπ to the initial solutions:
[tex]\theta=\dfrac{\pi}{3}+2k\pi,\;\;\theta=\dfrac{5\pi}{3}+2k\pi[/tex]
[tex]\dotfill[/tex]
Question 2
Given trigonometric equation:
[tex]3 \sin^2\theta - 7 \sin\theta +2=0[/tex]
Since the left side of the equation is a quadratic in the form 3a² - 7a + 2, where a = sin θ, we can factor it:
[tex]3 \sin^2\theta - 6 \sin\theta -\sin\theta +2=0\\\\3 \sin \theta(\sin\theta - 2) -1(\sin\theta -2)=0\\\\(3\sin\theta-1)(\sin\theta-2)=0[/tex]
Now, find the solutions by setting each factor equal to zero and solving for θ.
[tex]\sin\theta-2=0\\\\\sin\theta=2[/tex]
The parent sine function has a range between -1 and 1, so for sin θ = 2, there are no solutions because there is no value of θ that satisfies that equation.
[tex]3\sin\theta-1=0\\\\3\sin\theta=1\\\\\sin\theta=\dfrac{1}{3}\\\\\theta=0.339836909...[/tex]
As the sine of an angle is positive in quadrants I and II, the other solution is:
[tex]\theta=\pi-0.33983690... = 2.80175574...[/tex]
As the sine function repeats itself every 2π radians, we add 2kπ to account for all possible repetitions, where k is any integer. Therefore, the solutions rounded to 2 decimal places are:
[tex]\theta=0.34+2k\pi,\;\;\theta= 2.80+2k\pi[/tex]
[tex]\dotfill[/tex]
Question 3
Given trigonometric equation:
[tex]\sin^2 \theta = 2 \sin \theta +3[/tex]
Rearrange the equation so that it equals zero:
[tex]\sin^2 \theta - 2 \sin \theta - 3 = 0[/tex]
Factor:
[tex]\sin^2 \theta +\sin \theta - 3 \sin \theta - 3 = 0\\\\\sin \theta(\sin \theta + 1)- 3 (\sin \theta + 1) = 0\\\\(\sin \theta - 3)(\sin \theta + 1) = 0[/tex]
Now, find the solutions by setting each factor equal to zero and solving for θ.
[tex]\sin\theta-3=0\\\\\sin\theta=3[/tex]
The parent sine function has a range between -1 and 1, so for sin θ = 3, there are no solutions because there is no value of θ that satisfies that equation.
[tex]\sin\theta+1=0\\\\\sin\theta=-1[/tex]
According to the unit circle, sin θ equals -1 when θ = 3π/2. Therefore:
[tex]\theta=\dfrac{3\pi}{2}[/tex]
As the sine function repeats itself every 2π radians, we add 2kπ to account for all possible repetitions, where k is any integer. Therefore, the solutions are:
[tex]\theta=\dfrac{3\pi}{2}+2k\pi[/tex]
[tex]\dotfill[/tex]
Question 4
Given trigonometric equation:
[tex]\cos \theta \sin \theta- 2 \cos \theta =0[/tex]
Factor out the common term cos θ:
[tex]\cos \theta( \sin \theta- 2 ) =0[/tex]
Now, find the solutions by setting each factor equal to zero and solving for θ.
[tex]\sin\theta-2=0\\\\\sin\theta=2[/tex]
The parent sine function has a range between -1 and 1, so for sin θ = 2, there are no solutions because there is no value of θ that satisfies that equation.
[tex]\cos \theta=0[/tex]
According to the unit circle, cos θ equals zero when θ = π/2 and θ = 3π/2. Therefore:
[tex]\theta=\dfrac{\pi}{2},\;\;\theta=\dfrac{3\pi}{2}[/tex]
As the cosine function repeats itself every 2π radians, we add 2kπ to account for all possible repetitions, where k is any integer. Therefore, the solutions are:
[tex]\theta=\dfrac{\pi}{2}+2k\pi,\;\;\theta=\dfrac{3\pi}{2}+2k\pi[/tex]
