phinah 25 Jul 2018, 15:30
With an equation of the form 2 `sin^2` x + sin 2x = 0, I need help in determining a first step to solving it.
Tried to go the route of (2 sin x + ....)(sin x + .... ) but saw immediately it is a no-go.
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With an equation of the form 2 `sin^2` x + sin 2x = 0, I need help in determining a first step to solving it.
Relevant page <a href="https://www.intmath.com/analytic-trigonometry/5-trigonometric-equations.php">5. Trigonometric Equations</a> What I've done so far Tried to go the route of (2 sin x + ....)(sin x + .... ) but saw immediately it is a no-go.
Murray 26 Jul 2018, 21:28
My hint is to remind yourself of the first formula on this page: Double Angle Formulas
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My hint is to remind yourself of the first formula on this page: <a href="https://www.intmath.com/analytic-trigonometry/3-double-angle-formulas.php">Double Angle Formulas</a>
phinah 21 Aug 2018, 15:44
Okay thanks for that hint.
I substituted `2 sin x cos x` for `sin 2x.`
So `2\ sin^2 x + 2 sin x cos x = 0;`
Factoring: `2 sin x (sin x + cos x) = 0;
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`sin x = 0` or `sin x + cos x = 0;`
For `sin x = 0:` `x = 0,` `pi,` `2pi`;
For `sin x + cos x = 0,` `sin x = - cos x:` `x = (3pi)/4.`
X
Okay thanks for that hint. I substituted `2 sin x cos x` for `sin 2x.` So `2\ sin^2 x + 2 sin x cos x = 0;` Factoring: `2 sin x (sin x + cos x) = 0; ` `sin x = 0` or `sin x + cos x = 0;` For `sin x = 0:` `x = 0,` `pi,` `2pi`; For `sin x + cos x = 0,` `sin x = - cos x:` `x = (3pi)/4.`
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