# Where are the x-intercepts for f(x) = 4 cos(2x - π) from x = 0 to x = 2π?

**Solution:**

Given: x-intercepts for f(x) = 4 cos(2x - π) from x = 0 to x = 2π

We have to find if f(x) = 0

Consider x = π/4

f(π/4) = 4 cos(2π/4 - π)

Taking the LCM

f(π/4) = 4 cos[(2π - 4π)/4]

f(π/4) = 4 cos (-2π)/4

f(π/4) = 4 cos (-π)/2

f(π/4) = 4(0)

f(π/4) = 0

Consider x = 3π/4

f(3π/4) = 4 cos(2 × 3π/4 - π) = 4 cos[3π/2 - π]

Taking the LCM

f(3π/4) = 4 cos[3π - 2π]/2

f(3π/4) = 4 cos π/2

f(3π/4) = 4(0)

f(3π/4) = 0

Consider x = 5π/4

f(5π/4) = 4 cos(2 × 5π/4 - π) = 4 cos[5π/2 - π]

Taking the LCM

f(5π/4) = 4 cos[(5π - 2π)/2]

f(5π/4) = 4 cos (3π)/2

f(5π/4) = 4(0)

f(5π/4) = 0

Consider x = 7π/4

f(7π/4) = 4 cos(2 × 7π/4 - π)

= 4 cos[7π/2 - π]

Taking the LCM

f(7π/4) = 4 cos[(7π - 2π)/2]

f(7π/4) = 4 cos (5π)/2

f(7π/4) = 4(0)

f(7π/4) = 0

Therefore, the x-intercepts are π/4, 3π/4, 5π/4 and 7π/4.

## Where are the x-intercepts for f(x) = 4 cos(2x - π) from x = 0 to x = 2π?

**Summary:**

The x-intercepts for f(x) = 4 cos(2x - π) from x = 0 to x = 2π are π/4, 3π/4, 5π/4 and 7π/4.

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