Question

$\huge\underline\mathfrak\green{Question\:-}$



If sinØ + 2cosØ = 1, find the value of 2sinØ - cosØ.


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Answer 1

Answer:

2 sin Ø - cos Ø = ± 2

Step-by-step explanation:

Given :

sin Ø + 2 cos Ø =  1

Squaring on both side :

sin² Ø + 4 cos² Ø + 4 sin Ø cos Ø   =  1

Now using sin² Ø = 1 - cos² Ø  &  cos² Ø = 1 - sin² Ø

1 - cos² Ø + 4 ( 1 - sin² Ø ) + 4 sin Ø cos Ø = 1

- cos² Ø + 4  -  4 sin² Ø + 4 sin Ø cos Ø = 1 - 1

- cos² Ø   -  4 sin² Ø + 4 sin Ø cos Ø + 4 = 0

4 sin² Ø + cos² Ø - 4 sin Ø cos Ø + 4 = 4

( 2 sin Ø - cos Ø )² = 4

2 sin Ø - cos Ø = ± 2

Hence we get answer .

Answer 2

Answer:-

$\bold{ 2sin \theta - cos \theta = \pm 2 }$

Explanation:-

Given:-

$sin \theta + 2cos \theta = 1$

To Find:-

value of $2sin \theta - 2cos \theta$

Solution:-

$sin \theta + 2cos \theta = 1$

Squaring on both sides

${( sin \theta + 2cos \theta)}^{2} = {1}^{2}$

${ sin}^{2} \theta + 4{cos }^{2} \theta + 4 cos \theta. sin \theta= 1$

$1-{ cos}^{2} \theta + 4(1-{sin}^{2} \theta )+ 4 cos \theta. sin \theta= 1$

$1- {cos}^{2} \theta + 4-4{sin}^{2} \theta + 4 cos \theta. sin \theta= 1$

${cos}^{2} \theta + 4{sin}^{2} \theta - 4 cos \theta. sin \theta -5 = -1$

${( 2sin \theta - cos \theta)}^{2} = - 1 + 5$

${( 2sin \theta - cos \theta)}^{2} = 4$

Taking Square root on both sides

$2sin \theta - cos \theta =\pm 2$

Hence,

$\bold{ 2sin \theta - cos \theta = \pm 2 }$

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