Question

Please solve 5 f
(Trigonometric Equations)
#salute u if u solved it

Answer 1

`\beta=45\degree\:\:or\:\:\beta=135\degree`

Explanation:

We want to solve
`\tan \beta \sec \beta=√(2)`, where
`0\le \beta \le360\degree`.

We rewrite in terms of sine and cosine.

`(\sin \beta)/(\cos \beta) \cdot (1)/(\cos \beta) =√(2)`

`(\sin \beta)/(\cos^2\beta)=√(2)`

Use the Pythagorean identity:
`\cos^2\beta=1-\sin^2\beta`.

`(\sin \beta)/(1-\sin^2\beta)=√(2)`

`\implies \sin \beta=√(2)(1-\sin^2\beta)`

`\implies \sin \beta=√(2)-√(2)\sin^2\beta`

`\implies √(2)\sin^2\beta+\sin \beta- √(2)=0`

This is a quadratic equation in
`\sin \beta`.

By the quadratic formula, we have:

`\sin \beta=\frac{-1\pm \sqrt{1^2-4(√(2))(-√(2) ) } }{2\cdot √(2) }`

`\sin \beta=(-1\pm √(1^2+4(2) ) )/(2\cdot √(2) )`

`\sin \beta=(-1\pm √(9) )/(2\cdot √(2) )`

`\sin \beta=(-1\pm3)/(2\cdot √(2) )`

`\sin \beta=(2)/(2\cdot √(2) )` or
`\sin \beta=(-4)/(2\cdot √(2) )`

`\sin \beta=(1)/(√(2) )` or
`\sin \beta=-(2)/(√(2) )`

`\sin \beta=(√(2))/(2)` or
`\sin \beta=-√(2)`

When
`\sin \beta=(√(2))/(2)` ,
`\beta=\sin ^(-1)((√(2) )/(2) )`

`\implies \beta=45\degree\:\:or\:\:\beta=135\degree` on the interval
`0\le \beta \le360\degree`.

When
`\sin \beta=-√(2)`,
`\beta` is not defined because
`-1\le \sin \beta \le1`