Question

Find theta if 2cos theta + cos square theta=theta​

Answer 1

Step-by-step explanation:

Greater than or equal to

2

−3

and less than or equal to 3

We have,

cos2θ+2cosθ=2cos

2

θ−1+2cosθ=2(cosθ+

2

1

)

2

−

2

3

Now, 2(cosθ+

2

1

)

2

≥0 for all θ

∴2(cosθ+

2

1

)

2

−

2

3

≥

2

−3

for all θ

⇒cos2θ+2cosθ≥

2

−3

for all θ

Also, maximum value of this expression is 3.

Answer 2

Question

Solve the equation to find theta :

$\rm :\longmapsto\:2cos\theta + {cos}^{2}\theta = 0$

$\: \: \: \: \: \: (a) \: \: \theta = 30\degree$

$\: \: \: \: \: \: (b) \: \: \theta = 90\degree$

$\: \: \: \: \: \: (c) \: \: \theta = 45\degree$

$\: \: \: \: \: \: (d) \: \: \theta = 60\degree$

$\blue{\large\underline{\sf{Solution-}}}$

Given Trigonometric equation is

$\rm :\longmapsto\:2cos\theta + {cos}^{2}\theta = 0$

can be rewritten as after taking common,

$\rm :\longmapsto\:cos\theta(2 + cos\theta) = 0$

$\rm\implies \:cos\theta = 0 \: \: or \: \: cos\theta + 2 = 0$

$\rm\implies \:cos\theta = cos90 \degree \: \: or \: \: cos\theta = - 2$

$\rm\implies \: \theta =90 \degree \: \: or \: \: cos\theta = - 2 \: \: \: \: \: \red{\{rejected \: as \: - 1 \leqslant cos\theta \leqslant 1 \}}$

$\\ \purple{\bf\implies \:\boxed{\tt{ \: \: \theta \: = \: 90 \: \degree \: }}}$

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$