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.
Rewrite the equation with cos x as the variable.
2 cos^2x -1 + 8 cosx +9 = 0
2cos^2x +8 cosx +8 = 0
cos^2x +4x +4 = 0
(cosx +2)^2 = 0
Since cosx cannot equal -2 (for real values of x), there are no real solutions. Quadratic equations that can be written as the square of a monomial factor have equal roots.
actually it should have been
2cos^2 x - 1 + 8cosx + 9 = 0
2cos^2 x + 8cosx + 8 = 0
cos^2 x + 4x + 4 = 0
(cosx+2)(cosx+2)=0
so we have shown that for cosx it has equal roots. (the factor cosx+2 appears twice)
for the second part,
cosx + 2 = 0
cosx = -2
but for all real x, -1 ≤ cosx ≤ 1
so cosx = -2 is outside that domain, and there are no real roots.
To solve the equation cos2x + 8cosx + 9 = 0 and show that it has equal roots, follow these steps:
Step 1: Let's use a substitution to simplify the equation. Let u = cos(x). We can then represent the equation as:
u^2 + 8u + 9 = 0
Step 2: Factorize the quadratic equation:
(u + 1)(u + 9) = 0
Step 3: Set each factor equal to zero and solve for u:
u + 1 = 0 or u + 9 = 0
u = -1 or u = -9
Step 4: Substitute back u = cos(x) into the equation:
cos(x) = -1 or cos(x) = -9
Step 5: Determine if these solutions are valid. Since the range of cosine function is -1 ≤ cos(x) ≤ 1, we can conclude that cos(x) = -9 has no real solutions.
Therefore, the only real solution is cos(x) = -1, which has equal roots.
Hence, the equation cos2x + 8cosx + 9 = 0 has equal roots for cos(x) = -1 but no real roots for cos(x) = -9.
To solve the equation cos2x + 8cosx + 9 = 0 and show that it has equal roots, we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Let's apply this to our equation: cos2x + 8cosx + 9 = 0.
First, we need to rewrite the equation in terms of cosx.
Using the identity cos2x = 2cos^2 x - 1, the equation becomes:
2cos^2 x - 1 + 8cosx + 9 = 0
Rearranging terms:
2cos^2 x + 8cosx + 8 = 0
Now, we can apply the quadratic formula:
cosx = (-8 ± √(8^2 - 4(2)(8))) / (2 * 2)
cosx = (-8 ± √(64 - 64)) / 4
cosx = (-8 ± √0) / 4
Since the discriminant (√(64 - 64)) is equal to zero, we only have one solution for cosx. Therefore, the roots of the equation are equal.
Now, let's move on to part b.
To show that there are no real roots for x, we need to determine if the equation has any values of cosx that are not within the range of -1 to 1.
Given that cosx is always between -1 and 1, we can conclude that the equation has no real roots for x.
Therefore, there are no real solutions to the equation cos2x + 8cosx + 9 = 0.