Trig Problem- solve equation. Check

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The equation 5sin²(x) - 3 = 0 simplifies to sin²(x) = 3/5. To solve for x, the square root of both sides yields sin(x) = ±√(3/5). The inverse sine function gives solutions of approximately 50.8° and 129.2° for the positive root, while the negative root provides -50.8° and -129.2°. Thus, the complete set of solutions within the interval -180° ≤ x ≤ 180° is x = 50.8°, 129.2°, -50.8°, and -129.2°.
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Homework Statement



Solve the equation 5sin2x-3=0, given values of x in the interval -180≤x≤180 correct to 1 decimal place.

Homework Equations


for sin: take inverse
180-θ
±360


The Attempt at a Solution


I rearranged the equation to
sin2=3/5
but I am not quite sure what to do since sine is squared. I think it won't affect answer.

Then I did the inverse which was 36.9
subtrated 180 from inverse; 180- 36.9 = 143.1
Then I did 360±36.9 and 360±143.1
where both of these didn't fit in the interval.

Is this correct: x=36.9° and 143.1°
 
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FlopperJr said:

Homework Statement



Solve the equation 5sin2x-3=0, given values of x in the interval -180≤x≤180 correct to 1 decimal place.

Homework Equations


for sin: take inverse
180-θ
±360


The Attempt at a Solution


I rearranged the equation to
sin2=3/5
but I'm not quite sure what to do since sine is squared. I think it won't affect answer.

Then I did the inverse which was 36.9
subtracted 180 from inverse; 180- 36.9 = 143.1
Then I did 360±36.9 and 360±143.1
where both of these didn't fit in the interval.

Is this correct: x=36.9° and 143.1°
What do you mean by " ... I'm not quite sure what to do since sine is squared. I think it won't affect answer." ?

Is the solution to u = 5/3 the same as the solution to u2 = 5/3 ? ... Of course not.
 
Aww. Alright. Then how would I do it if sin is squared? Sin^2=3/5
 
FlopperJr said:
Aww. Alright. Then how would I do it if sin is squared? Sin^2=3/5
That should be sin2(x) = 3/5.

Take the square root of both sides of the above equation just like you would to solve u2 = 3/5 .

Don't forget the ± .

After that use the arc-sine function, sin-1 .

Alternatively, you can use the identity, cos(2x) = 1 - 2sin2(x).
 
Ohhh. So it would be sinx= +/- the sqr root 3/5. Which then the inverse would be 50.8. Then 180 minus 50.8 is 129.2. After that I would take those two answers and +/- 360 which doesn't work. So my two answers I got are 50.8 and 129.2.
 
FlopperJr said:
Ohhh. So it would be sinx= +/- the sqr root 3/5. Which then the inverse would be 50.8. Then 180 minus 50.8 is 129.2. After that I would take those two answers and +/- 360 which doesn't work. So my two answers I got are 50.8 and 129.2.
Those two answers are fine. They're for \displaystyle \sin(x)=\sqrt{\frac{3}{5}}\,.

Now find the solution for the ' - ' sign: \displaystyle \sin(x)=-\sqrt{\frac{3}{5}}\,.
 
Okay, for negative square root:
The inverse is -50.8
180-(-50.8). Is 230.8 which doesn't fit the range
But if I subtract from 360 I get -129.2
So my my two answers for this is -50.8 and -129.2 giving me a total of four answers
 
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