All categories

2 years ago

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

Mathematics

1 answer:

erica [24]2 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>

You might be interested in

Hhhhhhhhhheeeeeeeeellllllllllllllppppppp

ELEN [110]

Answer:

Since the base is the same, we can just multiply the exponents, and when we do so we get y^6.

7 0

2 years ago

Read 2 more answers

in the six grade 54% of the are boys. there are 189 boys in the sixth grade. what is the total number of the students in sixth g

notka56 [123]

<h2>189 boys is 54%</h2><h2>x is 46%(100-54)</h2><h2>Cross multiply:</h2><h2>189*46%=54%*x</h2><h2>86.94=0.54x</h2><h2>x=161</h2><h2>189 boys 161 girls</h2><h2>350 altogether</h2><h2>Hope this helps plz hit the crown :D</h2>

6 0

2 years ago

Read 2 more answers

Please help its number 4

Veseljchak [2.6K]

Answer:

E 60+60+60=180 sum of triangle's angels must be 180

6 0

2 years ago

Read 2 more answers

G varies inversely as the cube of h.

Arada [10]

Answer:

g = $\frac{k}{h^3}$

Step-by-step explanation:

Given that g varies inversely as h³ then the equation relating them is

g = $\frac{k}{h^3}$ ← k is the constant of variation

5 0

2 years ago

Which of the following are solutions to the equation below?

scoundrel [369]

Answer:

The correct options for the solution values are:

$x=2\sqrt{2}-5$

$x=-2\sqrt{2}-5$

Step-by-step explanation:

Given the expression

$x^2+10x+25=8$

Subtract 25 from both sides

$x^2+10x+25-25=8-25$

Simplify

$x^2+10x=-17$

Add 25 or 5² to both sides

$x^2+10x+5^2=8$

$x^2+10x+5^2=\left(x+5\right)^2$

so the expression becomes

$\left(x+5\right)^2=8$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x+5=\sqrt{8}$

Subtract 5 from both sides

$x+5-5=2\sqrt{2}-5$

$x=2\sqrt{2}-5$

solve

$x+5=-\sqrt{8}$

Subtract 5 from both sides

$x+5-5=-2\sqrt{2}-5$

$x=-2\sqrt{2}-5$

Therefore, the solution to the equation

$x=2\sqrt{2}-5,\:x=-2\sqrt{2}-5$

Hence, the correct options for the solution values are:

$x=2\sqrt{2}-5$

$x=-2\sqrt{2}-5$

6 0

2 years ago