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KIM [24]
4 years ago
7

What is the solution to 2sin^(2)x+sinx+1=0

Mathematics
1 answer:
Alexus [3.1K]4 years ago
3 0
Well you can try rewriting it to this

answer is   d  270

first start of by factoring  and subtracting the 1 into the right side

sin(x)  ( 2 sin (x)  + 1)  = -1  

set each one equal to -1 

sin( x) = -1   and   2 sin (x) +1 =  -1 

                              2 sin (x) = -2
                               sin ( x) = -1 
so therefore we have our final equation

sin ( x )  = - 1  and sin (x) = -1 

so then you look in your unit circle and find what coordinate equals -1 in terms of sin x 

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If , find the value of 2 square root of 17-x = 2x-10, find the value of x/4.
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2 \sqrt{17-x} =2x-10\ \ \ \ \Rightarrow\ \ \ D:(17-x \geq 0\ \ \ and\ \ \ 2x-10 \geq 0)\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq 17\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ x \geq 5\\\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=\\\\2 \sqrt{17-x} =2(x-5)\ /:2\\\\ \sqrt{17-x} =x-5\ \ \ \Rightarrow\ \ \ (\sqrt{17-x})^2 =(x-5)^2\\\\17-x=x^2-10x+25\\\\

x^2-9x+18=0\\\\ x^2-3x-6x+18=0\\\\x(x-3)-6(x-3)=0\\\\(x-3)(x-6)=0\ \ \ \ \Leftrightarrow\ \ \ (x-3=0\ \ \ \ or\ \ \ \ x-6=0)\\\\x=3\ \notin\ D\ \ \ \ \ \ \ \ \ \ \ x=6\ \in\ D\ \ \ \Rightarrow\ \ \  \frac{x}{4} = \frac{6}{4} =1.5\\\\Ans.\  \frac{x}{4} =1.5
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