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Popular Trigonometry >

(sin(x)+sin^2(x))/2 =0.5

  • Pre Algebra
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Solution

2sin(x)+sin2(x)​=0.5

Solution

x=0.66623…+2πn,x=π−0.66623…+2πn
+1
Degrees
x=38.17270…∘+360∘n,x=141.82729…∘+360∘n
Solution steps
2sin(x)+sin2(x)​=0.5
Solve by substitution
2sin(x)+sin2(x)​=0.5
Let: sin(x)=u2u+u2​=0.5
2u+u2​=0.5:u=2−1+5​​,u=−21+5​​
2u+u2​=0.5
Multiply both sides by 10
2u+u2​=0.5
To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by 102u+u2​⋅10=0.5⋅10
Refine5(u2+u)=5
5(u2+u)=5
Expand 5(u2+u):5u2+5u
5(u2+u)
Apply the distributive law: a(b+c)=ab+aca=5,b=u2,c=u=5u2+5u
5u2+5u=5
Move 5to the left side
5u2+5u=5
Subtract 5 from both sides5u2+5u−5=5−5
Simplify5u2+5u−5=0
5u2+5u−5=0
Solve with the quadratic formula
5u2+5u−5=0
Quadratic Equation Formula:
For a=5,b=5,c=−5u1,2​=2⋅5−5±52−4⋅5(−5)​​
u1,2​=2⋅5−5±52−4⋅5(−5)​​
52−4⋅5(−5)​=55​
52−4⋅5(−5)​
Apply rule −(−a)=a=52+4⋅5⋅5​
Multiply the numbers: 4⋅5⋅5=100=52+100​
52=25=25+100​
Add the numbers: 25+100=125=125​
Prime factorization of 125:53
125
125divides by 5125=25⋅5=5⋅25
25divides by 525=5⋅5=5⋅5⋅5
5 is a prime number, therefore no further factorization is possible=5⋅5⋅5
=53
=53​
Apply exponent rule: ab+c=ab⋅ac=52⋅5​
Apply radical rule: =5​52​
Apply radical rule: 52​=5=55​
u1,2​=2⋅5−5±55​​
Separate the solutionsu1​=2⋅5−5+55​​,u2​=2⋅5−5−55​​
u=2⋅5−5+55​​:2−1+5​​
2⋅5−5+55​​
Multiply the numbers: 2⋅5=10=10−5+55​​
Factor −5+55​:5(−1+5​)
−5+55​
Rewrite as=−5⋅1+55​
Factor out common term 5=5(−1+5​)
=105(−1+5​)​
Cancel the common factor: 5=2−1+5​​
u=2⋅5−5−55​​:−21+5​​
2⋅5−5−55​​
Multiply the numbers: 2⋅5=10=10−5−55​​
Factor −5−55​:−5(1+5​)
−5−55​
Rewrite as=−5⋅1−55​
Factor out common term 5=−5(1+5​)
=−105(1+5​)​
Cancel the common factor: 5=−21+5​​
The solutions to the quadratic equation are:u=2−1+5​​,u=−21+5​​
Substitute back u=sin(x)sin(x)=2−1+5​​,sin(x)=−21+5​​
sin(x)=2−1+5​​,sin(x)=−21+5​​
sin(x)=2−1+5​​:x=arcsin(2−1+5​​)+2πn,x=π−arcsin(2−1+5​​)+2πn
sin(x)=2−1+5​​
Apply trig inverse properties
sin(x)=2−1+5​​
General solutions for sin(x)=2−1+5​​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(2−1+5​​)+2πn,x=π−arcsin(2−1+5​​)+2πn
x=arcsin(2−1+5​​)+2πn,x=π−arcsin(2−1+5​​)+2πn
sin(x)=−21+5​​:No Solution
sin(x)=−21+5​​
−1≤sin(x)≤1NoSolution
Combine all the solutionsx=arcsin(2−1+5​​)+2πn,x=π−arcsin(2−1+5​​)+2πn
Show solutions in decimal formx=0.66623…+2πn,x=π−0.66623…+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for (sin(x)+sin^2(x))/2 =0.5 ?

    The general solution for (sin(x)+sin^2(x))/2 =0.5 is x=0.66623…+2pin,x=pi-0.66623…+2pin
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