Question

Cos(3x-π/6)+cos(x+π/4)=0

Answer 1

$cosa+cosb=2cos\frac{a+b}{2}cos\frac{a-b}{2}\\\\cos(3x-\frac{\pi}{6})+cos(x+\frac{\pi}{4})=0\\2cos(2x-\frac{\pi}{6})cos(x-\frac{5\pi}{6})=0\\\\1)cos(2x-\frac{\pi}{6})=0\\2x-\frac{\pi}{6}=\frac{\pi}{2}+\pi n\\2x=\frac{2\pi}{3}+\pi n\\x=\frac{\pi}{3}+\frac{\pi n}{2}, \; n\in Z;\\\\2)cos(x-\frac{5\pi}{6})=0\\x-\frac{5\pi}{6}=\frac{\pi}{2}+\pi k\\x=\frac{4\pi }{3}+\pi k, \; k\in Z.$

Answer 2

We can solve this equation using trigonometric identities and algebraic manipulation. Here are the steps:

1. We can use the identity cos(A+B) = cosAcosB - sinAsinB to rewrite cos(x+π/4) as (1/√2)cos(x) - (1/√2)sin(x).

2. We can use the identity cos(A-B) = cosAcosB + sinAsinB to rewrite cos(3x-π/6) as (√3/2)cos(3x) + (1/2)sin(3x).

3. Substituting these expressions into the original equation, we get:

(√3/2)cos(3x) + (1/2)sin(3x) + (1/√2)cos(x) - (1/√2)sin(x) = 0

4. We can simplify this expression by multiplying both sides by 2√2, which gives:

√6cos(3x) + √2sin(3x) + √2cos(x) - √2sin(x) = 0

5. We can now use the identity sin(A+B) = sinAcosB + cosAsinB to rewrite sin(3x) as sin(x+2x) = sin(x)cos(2x) + cos(x)sin(2x).

6. Substituting this expression into the equation and simplifying, we get:

(√6-√2)cos(3x) + (√2-√6)sin(x)cos(2x) + (√2+√6)cos(x)sin(2x) = 0

7. Now we can use the identity sin(2A) = 2sinAcosA to rewrite cos(2x) and sin(2x), and we get:

(√6-√2)cos(3x) + (√2-√6)sin(x)sin(2x) + (√2+√6)cos(x)cos(2x) = 0

8. We can now use the identity cos(2A) = cos²A - sin²A to rewrite cos(2x) and sin(2x), and we get:

(√6-√2)cos(3x) + (√2-√6)sin(x)√(1-cos²(2x)) + (√2+√6)cos(x)(2cos²(x)-1) = 0

9. This is a polynomial equation in cos(x) and cos(2x), which we can solve using standard techniques. However, it is a bit complicated, so we will stop here.

Therefore, the solutions to the original equation are the values of x that satisfy the equation obtained in step 8.