Ask question

Ask question

All categories

3 years ago

13

Solve the equation on the interval [0, 2pi) 2 cos^2 (theta/2) - 3 cos theta = 0

1 answer:

Ksivusya [100]3 years ago

3 0

Hello :
by idetity : 1+cos(2a) = 2cos²(a)
let : a = <span>θ/2
1+cos(2×</span>θ/2)=2cos²(θ/2)
1+cos(θ) = 2cos²(θ/2)
solve : 2 cos^2 (θ/2) - 3 cos(θ) = 0 ......(1)
(1) : 1+cos(θ)-3 cos(θ) =0
2cos(θ) = 1

cos(θ) = 1/2
in : [0, 2π<span>[
</span>( θ = π/3) or (θ = 5π/3)

You might be interested in

Which equation is NOT true?

MrRissso [65]

I'm gonna say c I'm not shore though

7 0

3 years ago

An ice cream cone has a radius of 7 cm and a height of 8 cm

zmey [24]

Answer:

TSA of ice cream cone = 330 cm^2

Step-by-step explanation:

Total Surface Area of ice cream cone = pie r( r+h )

22/7 * 7 (7+8)

22 ( 7+8)

22*15

= 330 cm. square

4 0

3 years ago

If peter piper picks pickled pepers, how many pepers has he picked?

Katena32 [7]

None. They're already pickled so someone else must've picked them before him.

4 0

3 years ago

The region bounded by y=(3x)^(1/2), y=3x-6, y=0

Ganezh [65]

Answer:

4.5 sq. units.

Step-by-step explanation:

The given curve is $y = (3x)^{\frac{1}{2} }$

⇒ $y^{2} = 3x$ ...... (1)

This curve passes through (0,0) point.

Now, the straight line is y = 3x - 6 ....... (2)

Now, solving (1) and (2) we get,

$y^{2} - y - 6 = 0$

⇒ (y - 3)(y + 2) = 0

⇒ y = 3 or y = -2

We will consider y = 3.

Now, y = 3x - 6 has zero at x = 2.

Therefor, the required are = $\int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx$

= $\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}$

= $[\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} } }{3}] - [13.5 - 18 - 6 + 12]$

= 6 - 1.5

= 4.5 sq. units. (Answer)

7 0

3 years ago

What is the slope of the line?

Eddi Din [679]

Answer:

-1

Step-by-step explanation:

Choose two point on that line such as (1,1) and (2,0)

Then use slope formula y-y/x-x

1-0/1-2 = 1/-1 = -1

8 0

3 years ago

Read 2 more answers