Question

If secXcos5X+1=0, where X is an acute angle, find value of X.

Answer 1

secXcos5X+1=0
or, secXcos5X=-1
or, cos5X=-1/secX
or, cos5X=-cosX
or, cos5X+cosX=0
or, 2cos(5X+X)/2cos(5X-X)/2=0
or, cos3Xcos2X=0
or, (4cos³X-3cosX)(2cos²X-1)=0
Either, 4cos³X-3cosX=0
or, 4cos³X=3cosX
or, cos²X=3/4
or, cosX=√3/2 (neglecting the negative sign)
∴, cosX=cos30° (∵, X is an acute angle)
or, x=30°
Or, 2cos²X-1=0
or, 2cos²X=1
or, cos²X=1/2
or, cosX=1/√2 (neglecting the negative sign)
∴, cosX=cos45° (∵, X is an acute angle)
or, X=45°
∴, X=30°, 45°

Answer 2

Step-by-step explanation:

Dear student,

We can solve this question as follows.

sec x . cos 5x + 1 = 0⇒cos 5xcos x = −1⇒cos 5x + cos x = 0⇒2 cos 3x . cos x = 0⇒cos 3x = 0 or cos x = 0⇒cos 3x = cos(π2) or cos x = cos(π2)⇒3x = π2 or x = π2⇒x = π6 or x = π2 [rejected]So, x = π6

Regards