The equation `cos 4theta+ sin 5 theta=2` is valid only when `cos 4 theta=1` and `sin 5 theta=1`. Thus, `4 theta=2npi and 5 theta=2m pi+pi//2, n , m in Z` `rArr theta=(2n pi)/4 and theta=(2m pi)/5+pi/10, n, m in Z` Putting n, `m=0, pm 1, pm 2`, ..., the common value in `[0, 2pi]` is `theta=pi//2`. Therefore, the solution is `theta=2k pi+pi//2, k in Z`.
Topper's Solved these Questions
TRIGONOMETRIC EQUATIONS
CENGAGE|Exercise Exercise 4.1|12 Videos
TRIGONOMETRIC EQUATIONS
CENGAGE|Exercise Exercise 4.2|6 Videos
TRIGNOMETRIC RATIOS IDENTITIES AND TRIGNOMETRIC EQUATIONS
CENGAGE|Exercise Question Bank|34 Videos
TRIGONOMETRIC FUNCTIONS
CENGAGE|Exercise SINGLE CORRECT ANSWER TYPE|38 Videos
Similar Questions
Explore conceptually related problems
Solve cos4 theta+sin5 theta=2
(i)If 3sin x+4cos ax=7 has atleast one solution; find the possible values of a.(ii) Solve cos^(50)x-sin^(50)=1( iii) Solve cos4 theta+sin5 theta=2
Solve : 3-2 cos theta -4 sin theta - cos 2theta+sin 2theta=0 .