Hello :
<span> sin 2x = sin x and 0 ≤ x ≤ 2π.
all solutions :
2x= x +2k</span>π or x= π -x +2kπ ..... k in : Z
x = 2kπ or : x = π/2 + kπ
but :
<span>0 ≤ x ≤ 2π
</span><span>all values of x such that sin 2x = sin x and 0 ≤ x ≤ 2π are :
</span>k = 0 : x=0 , x= π/2
k=1 : x=2 π , x=3π/2
Answer:
https://revisionmaths.com/gcse-maths-revision/shape-and-space/vectors
i find this helpful- hope u do too (can u mark my answer as brainliest?)
Answer:
so can't see the graph
Step-by-step explanation:
but if you can give a picture of the graph i could help out.
Well 6^2 is 36 so my best estimate is 6
Answer:
The complex number in the form of a + b i is 3/2 + i √3/2
Step-by-step explanation:
* Lets revise the complex number in Cartesian form and polar form
- The complex number in the Cartesian form is a + bi
-The complex number in the polar form is r(cosФ + i sinФ)
* Lets revise how we can find one from the other
- r² = a² + b²
- tanФ = b/a
* Now lets solve the problem
∵ z = 3(cos 60° + i sin 60°)
∴ r = 3 and Ф = 60°
∵ cos 60° = 1/2
∵ sin 60 = √3/2
- Substitute these values in z
∴ z = 3(1/2 + i √3/2) ⇒ open the bracket
∴ z = 3/2 + i √3/2
* The complex number in the form of a + b i is 3/2 + i √3/2
