Question

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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

EVALUATION

$10 \: {sin}^{2} \theta - 5\: {cos}^{2} \theta + 2 = 4\: {sin} \theta$

$\implies \: 10 \: {sin}^{2} \theta - 5(1 - \: {sin}^{2} \theta )+ 2 = 4\: {sin} \theta$

$\implies \: 10 \: {sin}^{2} \theta - 5+ 5 \: {sin}^{2} \theta + 2 = 4\: {sin} \theta$

$\implies \: 15 \: {sin}^{2} \theta - 4\: {sin} \theta - 3 = 0$

$\implies \: 15 \: {sin}^{2} \theta - 9\: {sin} \theta + 5\: {sin} \theta - 3 = 0$

$\implies \: 3 \: {sin} \theta\:( 5 {sin} \theta - 3) + 1( 5\: {sin} \theta - 3) = 0$

$\implies \:( 3 \: {sin} \theta\: + 1)( 5 {sin} \theta - 3) = 0$

$so \: either \: \:( 3 \: {sin} \theta\: + 1) = 0 \: \: or \: \: ( 5 {sin} \theta - 3) = 0$

$\displaystyle \: so \: either \: \: \: {sin} \theta\: = - \frac{ 1 }{3} \: \: or \: \: {sin} \theta = \frac{3}{5}$

If

$\displaystyle \: \: \: \: {sin} \theta\: = - \frac{ 1 }{3} \:$

Then x is in third or Fourth Quadrant

So in this case

$x \: = {199.5}^{ \circ} \: \: \: or \: \: {340.5}^{ \circ}$

If

$\displaystyle \: \: \: \: {sin} \theta\: = \frac{ 3 }{5} \:$

Then x is First or second quadrant

So in this case

$x \: = {36.9}^{ \circ} \: \: \: or \: \: {143.1}^{ \circ}$

RESULT

SO THE REQUIRED SOLUTION IS

$x \: = {36.9}^{ \circ} \: , \: {143.1}^{ \circ} \: ,{199.5}^{ \circ} \:, {340.5}^{ \circ}$

Answer 2

Answer:

10sin

2

θ−5cos

2

θ+2=4sinθ

\implies \: 10 \: {sin}^{2} \theta - 5(1 - \: {sin}^{2} \theta )+ 2 = 4\: {sin} \theta⟹10sin

2

θ−5(1−sin

2

θ)+2=4sinθ

\implies \: 10 \: {sin}^{2} \theta - 5+ 5 \: {sin}^{2} \theta + 2 = 4\: {sin} \theta⟹10sin

2

θ−5+5sin

2

θ+2=4sinθ

\implies \: 15 \: {sin}^{2} \theta - 4\: {sin} \theta - 3 = 0⟹15sin

2

θ−4sinθ−3=0

\implies \: 15 \: {sin}^{2} \theta - 9\: {sin} \theta + 5\: {sin} \theta - 3 = 0⟹15sin

2

θ−9sinθ+5sinθ−3=0

\implies \: 3 \: {sin} \theta\:( 5 {sin} \theta - 3) + 1( 5\: {sin} \theta - 3) = 0⟹3sinθ(5sinθ−3)+1(5sinθ−3)=0

\implies \:( 3 \: {sin} \theta\: + 1)( 5 {sin} \theta - 3) = 0⟹(3sinθ+1)(5sinθ−3)=0

so \: either \: \:( 3 \: {sin} \theta\: + 1) = 0 \: \: or \: \: ( 5 {sin} \theta - 3) = 0soeither(3sinθ+1)=0or(5sinθ−3)=0

\displaystyle \: so \: either \: \: \: {sin} \theta\: = - \frac{ 1 }{3} \: \: or \: \: {sin} \theta = \frac{3}{5}soeithersinθ=−

3

1

orsinθ=

5

3

If

\displaystyle \: \: \: \: {sin} \theta\: = - \frac{ 1 }{3} \:sinθ=−

3

1

Then x is in third or Fourth Quadrant

So in this case

x \: = {199.5}^{ \circ} \: \: \: or \: \: {340.5}^{ \circ}x=199.5

∘

or340.5

∘

If

\displaystyle \: \: \: \: {sin} \theta\: = \frac{ 3 }{5} \:sinθ=

5

3

Then x is First or second quadrant

So in this case

x \: = {36.9}^{ \circ} \: \: \: or \: \: {143.1}^{ \circ}x=36.9

∘

or143.1

∘

RESULT

SO THE REQUIRED SOLUTION IS

x \: = {36.9}^{ \circ} \: , \: {143.1}^{ \circ} \: ,{199.5}^{ \circ} \:, {340.5}^{ \circ}x=36.9

∘

,143.1

∘

,199.5

∘

,340.5

∘