q1) 2cos2x is f(x)
-square root3 is g(x)
solve f(x)-g(x)=0 to get points of intersection in x is between 0 and 180degrees
q2) a) write the equation cos2x + 8cosx+9=0 in terms of cosx and show that for cosx it has equal roots
q2b) show that there are no real roots for x.
for q2 i have tried to do it but i get upto the bit 2(cosx+2)(2cosx+2)=0 and i don't know what to do next.
could you please do all of question 1 as i don't get it at all and i feel if i see all then working then only then will i understand the method
hi there its me agfain MRMATHHEWMATICS!!!
i am sorry but i meant to write for question1 that 2cos2x is is g(x)
and that y=-squareroot 3 is the f(x)
#1
So you want to solve f(x)-g(x)=0 ?
-squareroot 3 - 2cos2x = 0
cos2x = -√3/2 , I just rearranged.
The cosine is negative in II and III, so
2x = 150º or 2x = 210º
x = 75º or x = 105º for 0 ≤ x ≤ 180º
Sure, let's go through both questions step by step.
Q1) To solve the equation f(x) - g(x) = 0, where f(x) = 2cos(2x) and g(x) = -√3, we need to find the points of intersection between the two functions within the range of x between 0 and 180 degrees.
Here's how you can solve it:
Step 1: Set the equation f(x) - g(x) = 0:
2cos(2x) - (-√3) = 0
Step 2: Simplify the equation:
2cos(2x) + √3 = 0
Step 3: Solve for cos(2x):
2cos(2x) = -√3
cos(2x) = -√3/2
Step 4: Find the possible angles for 2x using the unit circle or trigonometric identity:
cos(2x) = -√3/2 implies 2x = 150 degrees or 210 degrees (as these are the angles where cos(θ) = -√3/2).
Step 5: Solve for x:
Divide both sides by 2:
x = 75 degrees or 105 degrees (since x = 2x).
Therefore, the points of intersection between f(x) and g(x) within the range of x between 0 and 180 degrees are x = 75 degrees and x = 105 degrees.
Now, let's move on to Q2.
Q2a) To write the equation cos(2x) + 8cos(x) + 9 = 0 in terms of cos(x) and show that it has equal roots, follow these steps:
Step 1: Let's substitute cos(2x) = 2cos^2(x) - 1 into the equation:
2cos^2(x) - 1 + 8cos(x) + 9 = 0
Step 2: Simplify the equation:
2cos^2(x) + 8cos(x) + 8 = 0
Step 3: Divide the equation by 2:
cos^2(x) + 4cos(x) + 4 = 0
Step 4: Rewrite the equation using factoring, since the equation is quadratic:
(cos(x) + 2)^2 = 0
Step 5: Take the square root of both sides:
cos(x) + 2 = 0
Step 6: Solve for cos(x):
cos(x) = -2
So, we have found that the equation has only one root, which is cos(x) = -2.
Q2b) Now, let's show that there are no real roots for x when cos(x) = -2.
Step 1: Recall that the domain of the cosine function is -1 ≤ cos(x) ≤ 1, which means that the range of cos(x) is -1 ≤ cos(x) ≤ 1.
Step 2: Since -2 is not within the range of cos(x), there are no real solutions for cos(x) = -2.
Therefore, the equation cos(2x) + 8cos(x) + 9 = 0 has no real roots for x.
I hope this explanation helps you better understand the methods used to solve these problems. Let me know if you have any further questions!