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Verizon [17]
3 years ago
9

Simplify the following in the form of a + b √c2√2-7√8+4√½+8√32​

Mathematics
1 answer:
Kryger [21]3 years ago
4 0

Answer:

undefined

Step-by-step explanation:

the square root of a negative number doesnt exist in the set of Real numbers

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Step-by-step explanation:

Search for questions & chapters

Class 11

>>Maths

>>Trigonometric Functions

>>Trigonometric Equations

>>Solve for x: sin x + sin 2x...

Question

Bookmark

Solve for x:sinx+sin2x+sin3x=cosx+cos2x+cos3x in the interval 0≤x≤2π.

Medium

Solution

verified

Verified by Toppr

We write the given equation as

(sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x

or 2sin2xcosx+sin2x=2cos2xcosx+cos2x

or sin2x(2cosx+1)=cos2x(2cosx+1)

or (sin2x−cos2x)(2cosx+1)=0

∴sin2x−cos2x=0 or 2cosx+1=0

If sin2x−cos2x=0, then tan2x=1,

Hence 2x=nπ+π/4

or x=(4n+1)

8

π

.(1)

If 2cosx+1=0, then cosx=−1/2 (2)

∴x=2nπ±

3

2π

or x=

3

6n±2

π

We seek values of x in the interval 0≤x≤2π

In this interval (1) gives

x=π/8,5π/8,9π/8,13π/8. (n=0,1,2,3)

and (2) gives x=2π/3,4π/3. (for n=0,1)

Thus we get the answer

x=π/8,5π/8,2π/3,9π/8,4π/3,13π/8.

4 0
2 years ago
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