How Do You Solve a Trigonometric Equation Involving Both Sine and Cosine?

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SUMMARY

The discussion centers on solving the trigonometric equation 0.15348 = 0.1415cosβ - 0.291sinβcosβ. The user attempts to manipulate the equation using trigonometric identities, particularly by squaring both sides and substituting cos²β with a variable t. The transformation leads to a new equation involving t, but the user expresses uncertainty about how to proceed with finding the value of t. The conversation highlights the complexity of solving equations that involve both sine and cosine functions.

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  • Understanding of trigonometric identities, specifically sin²x + cos²x = 1
  • Familiarity with algebraic manipulation of equations
  • Knowledge of substitution methods in solving equations
  • Basic skills in handling square roots and quadratic equations
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  • Learn about the quadratic formula for solving equations in the form of at² + bt + c = 0
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as part of a physics problem i get to one stage before the ansewr and then i get stuck here:

0.15348=0.1415cosβ -0.291sinβcosβ

how do i solve this equation with both sinβ and cosβ, i realize that i need to play with the identities but have had no luck,
please help
 
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Try squaring both sides and see if you can use sin2x+cos2x=1 in hopes to get one trig term.
 
copied it all wrong, sorry, obviously can't work,,,

meant to be

0.15348=0.1415cosβ -0.291sinβcosβ

if i square it i get

0.023556=0.02cos2β +0.085sin2βcos2β-0.08sinβcosβ

if i say cos2β=t

0.023556=0.02t + 0.085(1-t)*(t)-0.08*\sqrt{1-t}\sqrt{t}

0.023556=0.02t + 0.085t - 0.085t2 -0.08\sqrt{t-t<sup>2</sup>}

0.023556=0.105t - 0.85t2 - 0.08\sqrt{t-t<sup>2</sup>}

now how would i find t??

you sure there's no better way?
 

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