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3 years ago

Solve on the interval [0,2pi): 2 sin^2 x-3 sin x + 1 = 0

Mathematics

1 answer:

erica [24]3 years ago

7 0

Answer:

How do you solve #2sin^2x+3sinx+1=0# and find all solutions in the interval #[0,2pi)#?

Trigonometry Trigonometric Identities and Equations Solving Trigonometric Equations

Answer:

Explanation:

We may identify this as a quadratic equation in #sinx# and factorise it as a trinomial to obtain the solutions as

<h3>#thereforesinx=-1/2 or sinx=-1#</h3>

<h3>#therefore x= (7pi)/6 or (11pi)/6 or (3pi)/2#</h3>

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Step-by-step explanation:

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2 years ago

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aleksklad [387]

9514 1404 393

Answer:

c) 16,500 m³

d) 277,088 mm³

a) V = LWH

b) V = πr²h

Step-by-step explanation:

The relevant volume formulas are ...

rectangular pyramid: V = 1/3LWH
cylinder: V = πr²h
rectangular prism: V = LWH

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14b. The volume of a cylinder is given by the formula ...

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Answer:

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Step-by-step explanation:

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