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zheka24 [161]
3 years ago
8

Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π). (Enter your answers as a comma-separated list

. Round your answers to three decimal places where appropriate.)
sin 2θ + sin θ = 0
Mathematics
1 answer:
lara31 [8.8K]3 years ago
8 0
Sin(2θ)+sin(<span>θ)=0
use double angle formula: sin(2</span>θ)=2sin(θ)cos(<span>θ).
=>
2sin(</span>θ)cos(θ)+sin(<span>θ)=0
factor out sin(</span><span>θ)
sin(</span>θ)(2cos(<span>θ)+1)=0
by the zero product property,
sin(</span>θ)=0 ...........(a) or
(2cos(<span>θ)+1)=0.....(b)
Solution to (a): </span>θ=k(π<span>)
solution to (b): </span>θ=(2k+1)(π)+/-(π<span>)/3
for k=integer

For [0,2</span>π<span>), this translates to:
{0, 2</span>π/3,π,4π/3}
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Answer: x>-6

Explanation: -2x -6>-18
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What is difference between -0.4 and 44.4 a) 40 b) 44 c) -44 d)44.8
ololo11 [35]

Answer: d

Step-by-step explanation:

44,4 - (-0,4) = 44,4 + 0,4= 44,8

3 0
3 years ago
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Help Q-Q 7/10c =4 1/5
shepuryov [24]

\huge\mathfrak\green{answer}

= q -   \frac{q7}{10} c =  \frac{41}{5}

\red{multiply}

= q -  \frac{cq7}{10} =  \frac{41}{5}

\red{subtract \: q \: from \: both \: sides}

= q -  \frac{cq7}{10}  - q =  \frac{41}{5} - q

\red{now \: simplify}

=  -  \frac{cq7}{10}  =  \frac{41}{5}  - q

\red{multiply \: 10 \: both \: sides}

= 10( -  \frac{cq7}{10} ) = 10. \frac{41}{5}  - 10q

\red{simplify}

=  - cq7 = 82 - 10q

\red{now \: divide}

=  \frac{ - cq7}{ - q7}  -  \frac{82}{ - q7}  -  \frac{10q}{ - q7}

\red{simplify}

= c =   - \frac{82 - 10q}{q7} (q≠0)

Brainliest? :) (I'd really appreciate if you mark me as brainliest)

\huge\mathfrak\green{thank \: you}

5 0
3 years ago
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Apply the distributive property to factor out the greatest common factor.
Tanzania [10]
I’m pretty sure it’s 86, because you need to subtract 2 from the first side then you have 86 = X
5 0
2 years ago
URGENT HELP PLEASE!
Vaselesa [24]

Answer:

(a) x=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{9\pi}{8},\dfrac{11\pi}{8}

(b) x=\dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}

Step-by-step explanation:

It is given that 0\leq x\leq 2\pi.

(a)

\sqrt{2}\sin 2x=1

\sin 2x=\dfrac{1}{\sqrt{2}}

\sin 2x=\dfrac{\pi}{4}

2x=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{9\pi}{4},\dfrac{11\pi}{4}     [\because \sin x=\sin y\Rightarrow x=n\pi+(-1)^ny]

x=\dfrac{\pi}{8},\dfrac{3\pi}{8},\dfrac{9\pi}{8},\dfrac{11\pi}{8}

(b)

\csc^2 x-\csc x-2=0

\csc^2 x-2\csc x+\csc x-2=0

\csc x(\csc x-2)+1(\csc x-2)=0

(\csc x+1)(\csc x-2)=0

\csc x=-1\text{ or }\csc x=2

\sin x=-1\text{ or }\sin x=\dfrac{1}{2}         [\because \sin x=\dfrac{1}{\csc x}]

x=\dfrac{3\pi}{2}\text{ or }x=\dfrac{\pi}{6},\dfrac{5\pi}{6}

Therefore, x=\dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}.

5 0
2 years ago
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