find the general value of theta(@)
1)root three cos @+sin@=root 2

2)sin ^2@-2cos@+0.25=0

3)root 2 sec @+tan @=1

4)root 3tan2@+root 3tan3@+tan2@tan3@=1

5)3(sec^2@+tan^2@)=5

Question

find the general value of theta(@)
1)root three cos @+sin@=root 2
2)sin ^2@-2cos@+0 25=0
3)root 2 sec @+tan @=1
4)root 3tan2@+root 3tan3@+tan2@tan3@=1
5)3(sec^2@+tan^2@)=5

Shridhar Kaushik · Accepted Answer

$(1) \sqrt{3} \cos θ + \sin θ = \sqrt{2} ........ (1)$

This is of the form

$a \cos θ + b \sin θ = c where a = \sqrt{3}, b = 1 and c = \sqrt{2} .$

Let a = r cos α and b = r sin α.

Then $\sqrt{3}$ = r cos α and 1 = r sin α

$\Rightarrow r = \sqrt{(\sqrt{3})^{2} + 1} = 2 and \tan α = \frac{1}{\sqrt{3}} \Rightarrow α = \frac{π}{6}$

From (1), we get –

$r \cos α \cos θ + r \sin α \sin θ = \sqrt{2} \Rightarrow r (\cos α \cos θ + \sin α \sin θ) = \sqrt{2} \Rightarrow r \cos (θ - α) = \sqrt{2} \Rightarrow 2 \times \cos (θ - α) = \sqrt{2} \Rightarrow \cos (θ - \frac{π}{6}) = \frac{1}{\sqrt{2}} [∵ r = 2 and α = \frac{π}{6}] \Rightarrow \cos (θ + \frac{π}{6}) = \cos \frac{π}{4} \Rightarrow θ - \frac{π}{6} = 2 n π \pm \frac{π}{4}; n \in Z \Rightarrow θ = 2 n π \pm \frac{π}{4} + \frac{π}{6} n \in Z \Rightarrow θ = 2 n π \pm \frac{π}{4} + \frac{π}{6} or θ = 2 n π - \frac{π}{4} + \frac{π}{6} \Rightarrow θ = 2 n π + \frac{5 π}{12} or θ = 2 n π - \frac{π}{12} Hence, θ = 2 n π + \frac{5 π}{12} or θ = 2 n π - \frac{π}{12} where n \in Z .$

$(2) \sin^{2} θ - 2 \cos θ + \frac{1}{4} = 0 \Rightarrow 1 - \cos^{2} θ - 2 \cos θ + \frac{1}{4} = 0 \Rightarrow \cos^{2} θ + 2 \cos θ - \frac{5}{4} = 0 \Rightarrow 4 \cos^{2} θ + 8 \cos θ - 5 = 0 \Rightarrow 4 \cos^{2} θ + 10 \cos θ - 2 \cos θ - 5 = 0 \Rightarrow 2 \cos θ (2 \cos θ + 5) - 1 (2 \cos θ + 5) = 0 \Rightarrow (2 \cos θ + 5) (2 \cos θ - 1) = 0 \Rightarrow 2 \cos θ + 5 = 0 or 2 \cos θ - 1 = 0 \Rightarrow \cos θ = \frac{- 5}{2} or \cos θ = \frac{1}{2} Since - 1 \leq \cos θ \leq 1. ∴ \cos θ = \frac{- 5}{2} is not possible. Thus, cos θ = \frac{1}{2} = \cos \frac{π}{3} \Rightarrow θ = 2 n π \pm \frac{π}{3}, n \in Z \Rightarrow θ = 2 n π + \frac{π}{3}, or 2 n π - \frac{π}{3}; n \in Z$

$(3) \sqrt{2} \sec θ + \tan θ = 1 \Rightarrow \frac{\sqrt{2}}{\cos θ} + \frac{\sin θ}{\cos θ} = 1 \Rightarrow \sqrt{2} + \sin θ = \cos θ \Rightarrow \cos θ - \sin θ = \sqrt{2} ......... (1)$

This is of the form a cos θ – b sin θ = c.

Where a = 1, b = 1, c = $\sqrt{2}$

Let a = r cos α and b = r sin α.

We have,

$r = \sqrt{2}, α = \frac{π}{4} [Same as 1^{st} question]$

From (1), We get –

$r (\cos θ \cos α - \sin θ \sin α) = \sqrt{2} \Rightarrow r \cos (θ + α) = \sqrt{2} \Rightarrow \sqrt{2} \cos (θ + \frac{π}{4}) = \sqrt{2} \Rightarrow \cos (θ + \frac{π}{4}) = 1 = \cos 0 ° \Rightarrow θ + \frac{π}{4} = 2 n π \pm 0, n \in Z \Rightarrow θ = 2 n π - \frac{π}{4} \forall n \in Z$

Due to paucity of time, it would not be possible for us to provide the answer to all your queries. Try the rest of questions yourself and if you face any difficulty then do get back to us.

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find the general value of theta(@)1)root three cos @+sin@=root 2 2)sin ^2@-2cos@+0.25=0 3)root 2 sec @+tan @=1 4)root 3tan2@+root 3tan3@+tan2@tan3@=1 5)3(sec^2@+tan^2@)=5

find the general value of theta(@)
1)root three cos @+sin@=root 2

2)sin ^2@-2cos@+0.25=0

3)root 2 sec @+tan @=1

4)root 3tan2@+root 3tan3@+tan2@tan3@=1

5)3(sec^2@+tan^2@)=5